Answer:4-nitrobenzoic acid is more acidic than Benzoic acid
4-chlorobenzoic acid is more acididc than 4-methyl benzoic acid
P-Nitrophenol is more acidic than meta-nitrophenol
Explanation:Acidity of an acid can be explained in terms of the stability of conjugate base formed.
1. 4-nitrobenzoic acid is more acidic as compared to benzoic acid because of the presence of nitro group at 4-position that is para position of the benzene ring. Nitro group is an electron withdrawing group and it withdraws the electron density through resonance effect.
Here the conjugate base would be benzoate anion which has a carboxylate anion attached with the benzene ring . So any group which can withdraw the electron density from benzoate anion will stabilise the benzoate anion and subsequently it would increase the acidity .
In case of benzoic acid there is no extra withdrawl of electron density whereas in case of 4-nitrobenzoic acid the nitro group stabilises the benzoate anion by withdrawing electron density thereby stabilising the benzoate anion and increasing the acidity.
2. 4-chlorobenzoic acid is more acidic than 4-methyl benzoic acid because 4-chlorobenzoic acid has Cl group which is a good electron withdrawing group through inductive effect so the benzoate anion formed can be stabilised by the electron withdrawing Cl atom which would increase the acidity of 4-chlorobenzoic acid.
4-methylbenzoic acid has an electron donating Methyl group which donates electron density through inductive effect hence a methyl group would intensify the negative charge on the benzoate anion through electron donation and subsequently it would destabilise the benzoate anion thereby decreasing its acidity.
3. In case of phenols the conjugate base formed is phenoxide anion and the negative charge that is its electron density is delocalised over the whole phenol ring. The negative charge electron density is more prominent at ortho and para position rather than the meta position. p-nitrophenol is more acidic than m-nitrophenol because p-nitrophenol has the nitro group at para position where it can stabilise the phenoxide anion through electron withdrawl via resonance whereas in case of m-nitrophenol as the nitro group is present at meta position so it can not stabilize prominently through electron withdrawl via resonance.
In each pair, 4-nitrobenzoic acid, 4-chlorobenzoic acid, and p-nitrophenol are more acidic due to their electron withdrawing groups that can better stabilize the negative charge after ionization.
Explanation:The acidity of a compound can be determined by its ability to donate a proton (hydrogen ion), and this ability is often influenced by other groups present in the compound. In each pair, the compound with the group that is more electron withdrawing will generally be more acidic because they can better stabilize the negative charge of the carboxylate ion after ionization occurs.
Benzoic acid vs. 4-nitrobenzoic acid: The presence of the nitro group in the 4-nitrobenzoic acid makes it more electron withdrawing compared to benzoic acid, making it more acidic.
4-Methylbenzoic acid vs. 4-chlorobenzoic acid: The chlorine atom in 4-chlorobenzoic acid is more electron withdrawing than the methyl group in 4-methylbenzoic acid, making 4-chlorobenzoic acid more acidic.
p-Nitrophenol vs. m-nitrophenol: In this case, p-nitrophenol would be more acidic. Although the nitro group is meta to the phenol group in m-nitrophenol, and para in p-nitrophenol, the para position allows for more effective resonance stabilization post-ionization, enhancing its acidity.
Learn more about acidity of compounds here:https://brainly.com/question/35095491
#SPJ6
The 1H NMR signal for bromoform (CHBr3) appears at 2065 Hz when recorded on a 300−MHz NMR spectrometer. If the spectrum was recorded on a 500−MHz instrument, what would be the chemical shift of the CHBr3 proton? Enter your answer in the provided box.
Answer:The chemical shift (δ) is 6.88ppm.
Explanation:
We have the following data :
Absorption frequency of the proton in bromoform=2065Hz
Frequency of the NMR spectrometer(instrument)=300MHz
The formula for calculating the chemical shift (δ) in PPM is:
chemical shift (δ)={[Frequency of proton(Hz) -Frequency of reference(Hz]÷Frequency of NMR spectrometer(MHz }
Using the formula for chemical shift we can calculate the value of chemical shift (δ) in ppm
chemical shift (δ) ={[2065Hz-0]÷300×10⁻⁶}
chemical shift (δ) =6.88×10⁶
chemical shift (δ)=6.88ppm for a 300MHz NMR spectrometer
The chemical shift (δ) is a ratio of frequency absorbed by proton with that of NMR spectrometer frequency hence the chemical shift value would remain same what ever NMR spectrometer frequency we use. Chemical shift basically tells us about the position of signal with respect to the reference compound of TMS(δ=0).
chemical shift (δ) is measured in
So the value of (δ) is same for any spectrometer used.
The chemical shift (δ) for a 500MHz NMR spectrometer used would also be 6.88PPM.
Alternatively since the frequency of proton absorbed is directly related to the magnetic field applied that is the frequency of NMR spectrometer hence:
Let the frequency of proton absorbed in 300MHZ=V₁=2065
Let the frequency of proton absorbed in 500MHZ=V₂=?
frequency of proton absorbed∝Applied magnetic field(Frequency of NMR spectrometer)
So V₁/V₂=[300×10⁶]/[500×10⁶]
V₂=[2065×500]÷300
V₂=3441HZ
For a 500 MHz proton the frequency of absorption would be 3441MHz
using this frequency we can calculate chemical shift (δ) using above formula:
(δ)=[3441-0]/[500×10⁻⁶]
(δ)=6.88PPM
Hence we obtain the same value of chemical shift in both the spectrometers.
This answer explains how to calculate the chemical shift of a nucleus using NMR spectroscopy when switching between different instrument frequencies.
Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful tool used by chemists to analyze the structure of molecules. In NMR spectroscopy, the chemical shift is a measure of the resonance frequency of a nucleus compared to a standard.
To find the chemical shift on a 500-MHz instrument, we use the formula: Chemical shift on New Instrument = (Resonance Frequency on Old Instrument) × (New Instrument Frequency) / (Old Instrument Frequency).
Plugging in the values, we get: Chemical shift = 2065 Hz × 500 MHz / 300 MHz = 3441.67 Hz.
Therefore, the chemical shift of the CHBr₃ proton on a 500-MHz NMR instrument would be 3441.67 Hz.
The four bonds of carbon tetrachloride (CCl4) are polar, but the molecule is nonpolar because the bond polarity is canceled by the symmetric tetrahedral shape. When other atoms substitute for some of the Cl atoms, the symmetry is broken and the molecule becomes polar. Rank the following molecules from the least polar to the most polar: CH2Br2, CF2Cl2, CH2F2, CH2Cl2, CBr4, CF2Br2.
Answer:The following would be the polarity order CH₂F₂>CF₂Br₂>CF₂Cl₂>CH₂Cl₂>CH₂Br₂>CBr₄.
Explanation:
The polarity of any bond is associated with its dipole moment.
Dipole moment is created when two charges having equal magnitude but opposite signs are separated by a distance. Dipole moment is a vector quantity and it has a direction.
Mathematically:
Dipole moment=Magnitude of charge×distance between the charges
The charge separation only occurs in a bond when the two atoms forming bond have different electronegativities .The atom having more electroegativity pulls the shared electron density between the bonds towards itself thereby generating partial charges on individual atoms .
The atom which pulls the shared electron density between the bonds (more electronegative atom) towards itself develops a partial negative charge and the atom (less electronegative atom) from which the electrondensity is pulled generates a partial positive charge.
This leads to the development of a dipole as partial charges with opposite signs on individual atoms are separated by the bond length and the direction of the dipole is towards the electron which withdraws the electron density.
So we can say that the molecules will only have polarity or the molecule would be polar when they have a net dipole moment.
Net dipole moment of molecule is the vector sum of individual dipole moments of individual bonds.
It is possible that individual bonds may create dipole moments but due to the overall symmetrical nature of molecule these individual bonds cancel each other and hence the net dipole moment of molecule is zero which means the molecule is non-polar.
The net dipole moment in case of CCl₄ is zero because the vector some of dipole moments created by individual bonds is zero which means the individual dipoles cancel each other leading to a net zero dipole moment.
This happens as CCl₄ has a tetrahedral structure which is very symmetric and hence the individual bonds in CCl₄ are polar on account of electronegativity difference between carbon and chlorine which leads to creation of individual dipoles but overall the net dipole moment of molecule is zero as these individual dipoles cancel each other and hence the molecular is non-polar.
So we can say that molecules would have high polarity if they have high dipole moment.
A molecule will only have high dipole moment when the charge separtion is more and magnitude of partial charges developed is also more. This would happen when there is greater electronegativity difference between the two bond forming atoms
So the following would be the polarity order of the given molecules:
CH₂F₂>CF₂Br₂>CF₂Cl₂>CH₂Cl₂>CH₂Br₂>CBr₄
CH₂F₂ has the highest electronegativity difference in between bonds(C-H&C-F) so they will generate more dipole moment and hence it would be most polar.
CF₂Br₂ has the 2nd highest electronegativity difference in between bonds(C-F&C-Br) so they will generate dipole moment and hence it would be polar but less than CH₂F₂ .
CF₂Cl₂ has the 3rd highest electronegativity difference in between bonds(C-F&C-Cl) so they will generate dipole moment and hence it would also be polar but less than CH₂F₂ &CF₂Br₂ .
CH₂Cl₂ has the 4th highest electronegativity difference in between bonds(C-H&C-Cl) so they will generate dipole moment and hence it would be polar but less than other 3 molecules .
CH₂Br₂ has the 5th highest electronegativity difference in between bonds(C-H&C-Br) so they will generate dipole moment and hence it would be the least polar of all.
CBr₄ would be non-polar in nature as the net dipole moment would be Zero.
The answer is CH₂F₂> CF₂Br₂> CF₂Cl₂> CH₂Cl₂> CH₂Br₂> CBr₄
Further ExplanationIn chemistry, polarity (or polarity) is the separation of electrical charges that leads to molecules or chemical groups that have dipole or multipole electric moments. Polar molecules must contain polar chemical bonds due to electronegativity differences between bonding atoms. Polar molecules with two or more polar bonds must have asymmetrical geometry so that the bonding moments do not cancel out. Polar molecules interact through the dipole-dipole intermolecular forces and hydrogen bonds. Polarity underlies several physical properties including surface tension, solubility, and melting and boiling points.
Polarity Classification
Bonds can be categorized as extreme - very nonpolar or very polar. Completely nonpolar bonds occur when electronegativity is identical and therefore has a zero difference. Fully polar bonds are more accurately called ionic bonds, and occur when the difference between electronegativity is large enough that one atom takes electrons from another. The terms "polar" and "nonpolar" are usually applied to covalent bonds, which are bonds where the polarity is incomplete. To determine the polarity of covalent bonds using numerical tools, the difference between atomic electronegativity is used.
The bond polarity is usually divided into three groups based on the difference in electronegativity between the two bonding atoms. According to the Pauling scale:
Nonpolar bonds generally occur when the electronegativity difference between the two atoms is less than 0.5 Polar bonds generally occur when the electronegativity difference between the two atoms is approximately between 0.5 and 2.0 Ionic bonding generally occurs when the electronegativity difference between two atoms is greater than 2.0Learn more
definition of Polarity https://brainly.com/question/12944064
Polarity Classification https://brainly.com/question/12944064
Details
Grade: College
Subject: Chemistry
keywords: Polarity
The limiting reactant determines what the actual yield is. (T/F)
Answer:
True
Explanation:
Limiting reactant - the reactant which get completely consumed in a chemical reaction , is known as the limiting reactant .
As, the concentration of limiting reactant after the completion of the reaction will be zero , hence, it is used to determine the concentration of other reactants .
For example,
for a general reaction -
A + B ---> 3C
Assuming B to be the limiting reactant ,
hence, the concentration of C and A can be determined as -
1 mol of B can give 3 mol of C and 1 mol of A is used for the reaction.
The density of water at 4ºC is 1.00 x 103 kg/m3. What is water's density at 82ºC? Assume that the water's coefficient of volume expansion is constant. Enter your answer in kg/m3, without units, to the nearest whole number.
Answer:
Density of water at [tex] T = 82^0 C , \rho = 983.308 kg/m^3 [/tex]
Explanation:
The relationship between density and temperature is shown below:
[tex] \rho_1 = \rho_0 [ 1- \beta \Delta T ][/tex]
Where,
[tex] \rho_1[/tex] is the density at temperature [tex]T_1[/tex]
[tex] \rho_0[/tex] is the density at temperature [tex]T_0[/tex]
[tex] \beta [/tex] is the coefficient of volume expansion
[tex] \Delta T [/tex] is the change in temperature which is:
[tex] \Delta T = {T_1} -{T_0} [/tex]
Given,
[tex]T_0 = 4^0 C [/tex]
[tex] \rho_0 = 1.00\times 10^3 kg/m^3 [/tex]
[tex]T_1 = 82^0 C [/tex]
[tex] \Delta T = (82 -4) ^0 C =78 ^0 C [/tex]
[tex] \rho_1 = ? [/tex]
Also,
[tex] \beta for water = 0.000214 ^0C^{-1} [/tex]
So,
[tex] \rho_1 is: [/tex]
[tex] \rho_1 = 1.00\times 10^3 kg/m^3[1 - 0.000214 ^0C^{-1} \times 78^0 C ][/tex]
[tex] \rho_1 = 983.308 kg/m^3 [/tex]
Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated with HCl requires 11.22 mL of 0.0983 M HCl to reach the end point. Part A Calculate Ksp for Ca(OH)2.
Answer: The [tex]K_{sp}[/tex] for calcium hydroxide is [tex]5.324\times 10^{-6}[/tex]
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M[/tex]
The concentration of [tex]Ca(OH)_2[/tex] comes out to be 0.011 M.
The balanced equilibrium reaction for the ionization of calcium hydroxide follows:
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]
The expression for solubility constant for this reaction follows:
[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]
Putting the values in above equation, we get:
[tex]K_{sp}=(0.011)\times (2\times 0.11)^2[/tex]
[tex]K_{sp}=5.324\times 10^{-6}[/tex]
Hence, the [tex]K_{sp}[/tex] for calcium hydroxide is [tex]5.324\times 10^{-6}[/tex]
Answer:
[tex]5.2*10^{-6}[/tex]
Explanation:
The balanced chemical equation of the reaction is :
Ca(OH)2 + 2HCl → CaCl2 + 2 H20.
Ksp can be calculated by the following formula:
Ksp = [Ca^{2+} ]+ [OH^{2-}].
Moles of HCl = Molarity × Volume of solution ( liters).
Moles of HCl can be calculated by multiplying 0.01122 (liters) ×0.0983
Moles of HCl = 0.0011 or [tex]1.0*10^{-3}[/tex]
The calculation of the concentration of Calcium hydroxide ( as starting with 50 ml) is :
[tex]Ca(OH)_2 = \frac{1/2 * 0.0011}{0.05 (liters)}[/tex]
[tex]Ca(OH)_2 =0.011[/tex].
[tex]Ca(OH)_2 = 1.1 \times 10^{-2}[/tex].
Ksp = [Ca^{2+} ]+ [2OH^{2-}].
Ksp = [tex]1.1 \times 10^{-2}* (2.2 \times 10^{-2})^2[/tex]
Ksp = [tex]5.2*10^{-6}[/tex]
Hence, the Ksp of calcium hydroxide is [tex]5.2*10^{-6}[/tex]
Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 13.7 °C are mixed. The specific heat of water is 4.184 J/g°C.
Answer:
The final temperature of the system is 42.46°C.
Explanation:
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))[/tex]
where,
c = specific heat of water= [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of water sample with 100 °C= 50.0 g
[tex]m_2[/tex] = mass of water sample with 13.7 °C= 100.0 g
[tex]T_f[/tex] = final temperature of system
[tex]T_1[/tex] = initial temperature of 50 g of water sample= [tex]100^oC[/tex]
[tex]T_2[/tex] = initial temperature of 100 g of water =[tex]13.7^oC[/tex]
Now put all the given values in the given formula, we get
[tex]50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))[/tex]
[tex]T_f=42.46^oC[/tex]
The final temperature of the system is 42.46°C.
The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.60 mol of ethane is burned in an excess of oxygen?
When 5.60 mol of ethane is burned, 11.20 mol of CO2 are produced.
Explanation:The balanced equation for the combustion of ethane (C2H6) is 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g). From this equation, we can see that for every 2 moles of ethane burned, 4 moles of CO2 are produced. Therefore, to calculate how many moles of CO2 are produced when 5.60 mol of ethane is burned, we can use the ratio:
2 moles of ethane : 4 moles of CO2 = 5.60 mol of ethane : x moles of CO2
Solving for x, we find that x = (5.60 mol of ethane) x (4 moles of CO2) / (2 moles of ethane) = 11.20 mol of CO2.
Learn more about Combustion of Ethane here:https://brainly.com/question/26891476
#SPJ3
5.60 moles of ethane produce 11.20 moles of carbon dioxide when burned in excess oxygen, based on the 2:4 molar ratio from the balanced combustion equation. This demonstrates the stoichiometric calculation for the reaction. Thus, burning ethane produces twice as many moles of CO₂.
To solve this, we need to use the balanced chemical equation for the combustion of ethane:
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)
This equation tells us that 2 moles of ethane produce 4 moles of carbon dioxide. Therefore, the ratio of moles of ethane to moles of carbon dioxide is 2:4, or 1:2.
If 5.60 moles of ethane (C₂H₆) are burned, we can use this ratio to find the moles of CO₂ produced:
Identify the molar ratio: 1 mole C₂H₆ → 2 moles CO₂Calculate the moles of CO₂: 5.60 moles C₂H₆ × (2 moles CO₂ / 1 mole C₂H₆) = 11.20 moles CO₂Thus, burning 5.60 moles of ethane produces 11.20 moles of carbon dioxide.
Water gas (which is is a mixture of H2 and CO) can be made by reacting red hot charcoal (C) with steam (H2O). a. Write the balanced equation for this reaction. b. Show, how can the obtained water gas itself be used as a fuel (write the balanced reactions of complete combustion of H2 and CO).
To balance chemical equations you add coefficients in front of each molecule in such a way that the number of atoms in reactants are equal with the number of the atoms in products.
a) C + H₂O → CO + H₂
the chemical equation is balanced, no need for other coefficients (it is assumed 1 in front of each molecule as a coefficient)
b) Combustion it means that the hydrogen and carbon monoxide will react with oxygen.
2 H₂ + O₂ → 2 H₂O
2 CO + O₂ → 2 CO₂
Answer: The equations are written below.
Explanation:
A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.
The chemical equation for the formation of water gas follows:
[tex]C+H_2O\rightarrow H_2+CO[/tex]
The gases released are hydrogen gas and carbon monoxide gas.
Combustion reactions are defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.
The chemical equation for the combustion of hydrogen gas follows:
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
The chemical equation for the combustion of carbon monoxide gas follows:
[tex]2CO+O_2\rightarrow 2CO_2[/tex]
Hence, the equations are written above.
Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Nitric acid HNO3 Calcium hydroxide Ca(OH)2 Acetic acid H3CCOOH Methyl amine CH3NH2 Potassium chloride KCl Ethanol C2H5OH Glucose C6H12O6
Answer: Strong electrolytes:[tex](HNO_{3})[/tex] Nitric acid, [tex](Ca(OH)_{2})[/tex])Calcium Hydroxide and (KCl) potassium chloride
Weak electrolytes: [tex](CH_{3}COOH)[/tex]Acetic acid and [tex](CH_{3}NH_{2})[/tex] Methyl amine
Non-electrolytes:[tex](C_{2}H_{5}OH)[/tex]Ethanol and [tex](C_{6}H_{12}O_{6})[/tex]Glucose
Explanation: Electrolytes are those compounds which can conduct electricity when dissolved in any polar solvent.
Strong electrolytes are those compounds which completely ionise when dissolved in polar solvent and hence produce ions in solution . So greater the capacity of an compound to ionize itself greater number of ions would be present in solution and hence greater will be the capacity of the solution to conduct electricity.
Ionic compounds like [tex](HNO_{3})[/tex] Nitric acid ,(KCl) Potassium chloride and [tex](Ca(OH)_{2})[/tex])Calcium hydroxide are completely ionized when dissolved in polar solvent so these compounds are strong electrolytes.
Weak electrolytes are those compounds which undergo partial ionization when dissolve in polar solvents . So they are not able to produce more ions in the solution and hence the conductivity of a solution containing weak electrolytes is low.
[tex](CH_{3}COOH)[/tex]Acetic acid and [tex]CH_{3}NH_{2}[/tex]Methyl amine are partially ionized when dissolved in polar solvent so these electrolytes are weak electrolytes.
Non-electrolytes are those compounds which are not at all ionized in the polar solvent and they remain as molecules itself even if they are dissolved.
[tex](C_{2}H_{5}OH)[/tex]Ethanol and [tex](C_{6}H_{12}O_{6})[/tex]Glucose do not ionize when dissolved in polar solvent and remain as molecules itself so the solutions of these compounds will not have ions and hence they would be unable to conduct electricity.
so
What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka(HCIO) 4.0 x 10-8] 0.032g ? 2.4 g 04.1 g 9.1 g 20. g
Answer : The mass of potassium hypochlorite is, 4.1 grams.
Explanation : Given,
pH = 10.20
Volume of water = [tex]4.50\times 10^2ml=0.45L[/tex]
The decomposition of KClO will be :
[tex]KClO\rightarrow K^++ClO^-[/tex]
Now the further reaction with water [tex](H_2O)[/tex] to give,
[tex]ClO^-+H_2O\rightarrow HClO+OH^-[/tex]
First we have to calculate the pOH.
[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8[/tex]
Now we have to calculate the [tex]OH^-[/tex] concentration.
[tex]pOH=-\log [OH^-][/tex]
[tex]3.8=-\log [OH^-][/tex]
[tex][OH^-]=1.58\times 10^{-4}M[/tex]
Now we have to calculate the base dissociation constant.
Formula used : [tex]K_b=\frac{K_w}{K_a}[/tex]
Now put all the given values in this formula, we get :
[tex]K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}[/tex]
Now we have to calculate the concentration of [tex]ClO^-[/tex].
The equilibrium constant expression of the reaction is:
[tex]K_b=\frac{[OH^-][HClO]}{[ClO^-]}[/tex]
As we know that, [tex][OH^-]=[HClO]=1.58\times 10^{-4}M[/tex]
[tex]2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}[/tex]
[tex][ClO^-]=0.0999M[/tex]
Now we have to calculate the moles of [tex]ClO^-[/tex].
[tex]\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole[/tex]
As we know that, the number of moles of [tex]ClO^-[/tex] are equal to the number of moles of KClO.
So, the number of moles of KClO = 0.0449 mole
Now we have to calculate the mass of KClO.
[tex]\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO[/tex]
[tex]\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g[/tex]
Therefore, the mass of potassium hypochlorite is, 4.1 grams.
To find the amount of potassium hypochlorite required, we first calculate the concentration of [OH-] ions from the given pH, then use this to calculate the amount of hypochlorite ions required. Our calculations yield an approximate amount of 0.15g which, among the provided options, the closest is 0.032g.
Explanation:In this problem, we are asked to determine the mass of potassium hypochlorite that must be added to water to give a solution with a pH of 10.20. Potassium hypochlorite is a weak base, and the formula for pH is pH = 14 - pOH. Since pOH is the negative log of the concentration of OH- ions, we can rearrange to find [OH-].
First, find the pOH: pOH = 14 - pH = 14 - 10.2 = 3.8. Then find the [OH-]: [OH-] = 10^-pOH = 10^-3.8. This gives the concentration of hypochlorite ions (OCl-) in solution because in water it dissociates as KOCl -> K+ + OCl-.
Using the molar mass of potassium hypochlorite, we can find the mass that must be added. The molar mass is given as 90.6 g/mol. So to find the mass, we multiply the volume of the water (which must be in liters, so 4.5 * 10^-2 L) by the [OH-] and then by the molar mass of potassium hypochlorite. Thus, mass = volume * [OH-] * molar mass =~ 0.15g. Hence the closest answer is 0.032g.
Learn more about Potassium Hypochlorite here:https://brainly.com/question/31559073
#SPJ6
An enzyme that follows Michaelis-Menten kinetics has a KM value of 15.0 μM and a kcat value of 221 s−1. At an initial enzyme concentration of 0.0100 μM, the initial reaction velocity was found to be 1.07×10−6 μM/s. What was the initial concentration of the substrate, [S], used in the reaction ?
The initial substrate concentration was approximately 7.29 μM.
To determine the initial substrate concentration, we use the Michaelis-Menten equation along with the provided data to calculate that the initial substrate concentration was approximately 7.29 μM.
To determine the initial concentration of the substrate, [S], used in the reaction, we can use the Michaelis-Menten equation, which is:
[tex]vo=\frac{V(max)[S]}{Km+[S]}[/tex]
Given the following data:
Initial reaction velocity (v0) = 1.07×10⁻⁶ μM/sMichaelis constant (KM) = 15.0 μMTurnover number (kcat) = 221 s⁻¹Initial enzyme concentration ([E]₀) = 0.0100 μMFirst, we need to calculate Vmax using the formula:
Vmax = kcat [E]₀
Substituting the given values:
Vmax = 221 s⁻¹ x 0.0100 μM = 2.21 μM/s
Next, substitute Vmax and the given values into the Michaelis-Menten equation to solve for [S]:
1.07×10⁻⁶ = (2.21 [S]) / (15.0 + [S])
Rearranging to solve for [S] gives:
[S] = (1.07×10⁻⁶ * (15.0 + [S])) / 2.21
Solving this equation numerically, we find:
[S] ≈ 7.29×10⁻³ μM
Thus, the initial concentration of the substrate, [S], used in the reaction was approximately 7.29 μM.
A 40.15 gram sample of a hydrate of Na2SO4 was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, 21.27 grams of the anhydrous compound remained. What is the formula of the hydrate?
Answer:
The formula of the hydrate of the sodium sulfate is :[tex]Na_2SO_4.7H_2O[/tex]
Explanation:
Mass of hydrated sodium sulfate = 40.15 gram
Mass of completely dehydrated sodium sulfate = 21.27 gram
Mass of water molecules present in hydrated sodium sulfate =x
40.15 gram = 21.27 gram + x (Law of conservation of mass)
x = 40.15 gram - 21.27 gram = 18.88 g
Moles of water = [tex] \frac{18.88 g}{18 g/mol}= 1.04mol[/tex]
Moles of sodium sulfate =[tex]\frac{21.27 g}{142.04 g/mol}=0.1497 mol[/tex]
Whole number ratio of sodium sulfate and water:
Sodium sulfate =[tex]\frac{0.1497 mol}{0.1497 mol}=1[/tex]
Water =[tex]\frac{1.04 mol}{0.1497 mol}=6.9 a\\rox 7[/tex]
The formula of hydrate be [tex]Na_2SO_4.7H_2O[/tex]
So, the formula of the hydrate of the sodium sulfate is :[tex]Na_2SO_4.7H_2O[/tex]
The formula of the hydrate, the mass of water lost is calculated by subtracting the mass of anhydrous Na₂SO₄ from the original hydrate mass. The number of moles of each substance is then calculated and the ratio of moles of water to moles of Na₂SO₄ is approximately 7, giving a hydrate formula of Na₂SO₄·7H₂O.
The formula of the hydrate of Na₂SO₄, we first determine the mass of the water lost during heating by subtracting the mass of the anhydrous Na₂SO₄ from the original mass of the hydrate:
Mass of water lost = original mass - mass of anhydrous Na₂SO₄
Mass of water lost = 40.15 grams - 21.27 grams = 18.88 grams
Now, to find the number of moles of Na₂SO₄ and H₂O, we use their molar masses (Na₂SO₄ = 142.04 g/mol, H₂O = 18.01 g/mol):
Moles of Na₂SO₄ = 21.27 grams / 142.04 g/mol = 0.1498 moles
Moles of H₂O = 18.88 grams / 18.01 g/mol = 1.048 moles
The mole ratio of H₂O to Na₂SO₄ is found by dividing the moles of H₂O by the moles of Na₂SO₄:
Mole ratio = moles of H₂O / moles of Na₂SO₄
Mole ratio = 1.048 moles / 0.1498 moles ≈ 7
Therefore, the empirical hydrated compound formula is Na₂SO₄·7H₂O.
How many iron atoms are in 0.32 mol of Fe2031? 3.9x 1023 jron atoms O 3.9 iron atoms O 6.02 x 1023 iron atoms 1.9 x 1023 iron atoms O 11x 10-24 iron atoms
Answer: [tex]3.9\times 10^{23}[/tex] iron atoms
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]1 molecule of [tex]Fe_2O_3[/tex] contains= 2 atoms of iron
[tex]1 mole of [tex]Fe_2O_3[/tex] contains=[tex]2\times 6.023\times 10^{23}=12.05\times 10^{23}[/tex] atoms of iron
thus 0.32 moles of [tex]Fe_2O_3[/tex] contains=[tex]\frac{12.05\times 10^{23}}{1}\times 0.32=3.9\times 10^{23}[/tex] atoms of iron
Thus the sample would have [tex]3.9\times 10^{23}[/tex] iron atoms.
The density of the solid phase of a substance is 0.90 g/cm3 and the density of the liquid phase is 1.0 g/cm3. A large increase in pressure will a. lower the freezing point b. raise the freezing point c. lower the boiling point d. raise the triple point e. lower the triple point
The correct answer is: b. raise the freezing point.
When pressure increases, the freezing point of a substance generally increases. This phenomenon is known as the colligative property of freezing point elevation. In a system where solid and liquid phases coexist, an increase in pressure tends to favor the denser phase.
Since the liquid phase is denser than the solid phase, an increase in pressure would promote the transition from solid to liquid, causing the freezing point to rise. This effect is commonly observed in many substances and is utilized in various applications, such as in the preservation of food through high-pressure processing.
Therefore, under increased pressure, the freezing point of the substance would be raised, making option b the correct choice.
Sodium nitrite (NaNO2) reacted with 2−iodooctane to give a mixture of two constitutionally isomeric compounds of molecular formula C8H17NO2 in a combined yield of 88%. Draw reasonable structures for these two isomers. Click the "draw structure" button to launch the drawing utility. Place the two compounds in the appropriate boxes below.
Hey there!:
the chemical reaction between 2-iodoctane and sodium nitrite is as follows:
Answer:
On the attached picture.
Explanation:
Hello,
In the case, the reaction between 2-iodooctane and sodium nitrite, leads to the formation of an alkyl nitrite and a nitro alkane as shown on the attached picture. Once the reaction began, the salt breaks and the sodium bonds with the iodine from the 2-iodooctane to form sodium iodide, in such a way, a free radical in the second carbon is formed so the NO₂ could bond both as a nitrite and as a nitro radical; therefore, the formed species are octyl 2-nitrite and 2-nitrooctane.
Best regards.
The following reaction was carried out in a 2.00 L reaction vessel at 1100 K: C(s)+H2O(g)⇌CO(g)+H2(g) If during the course of the reaction, the vessel is found to contain 8.75 mol of C, 12.9 mol of H2O, 4.00 mol of CO, and 7.20 mol of H2, what is the reaction quotient Q?
The ratio of products to reactants in a process at a particular time is measured by the reaction quotient Q. The reaction quotient Q is 0.2552.
What is the reaction quotient ?The reactivity factor Q measures the proportional amounts of reactants and products present in a reaction at a specific time. A reaction's direction of shift toward equilibrium can be predicted using Q.
A reaction will go forward and change reactants into products if K > Q. The reaction will go in the opposite direction, turning reactants into products, if K Q. The system is already in equilibrium if Q = K.
Volume of container = 2 L
Concentration = mass / volume
Moles of carbon = 8.75 mol
Concentration of carbon = 8.75 / 2
= 4.375 mol
Moles of H₂O = 12.9 mol
Concentration of water = 12.9 /2
= 6.45 mol
Moles of CO = 4
Concentration of CO = 4 / 2
= 2
Moles of Hydrogen = 7.20 mol
Concentration of Hydrogen is 7.20 / 2
= 3.6 mol
Q is given as
Q = ( H₂ ) ( O₂ ) / ( C ) ( H₂O )
= 7.20 / 28.21
= 0.2552
Thus, The reaction quotient Q is 0.2552.
To learn more about reaction quotient, follow the link;
https://brainly.com/question/24202150
#SPJ2
Final answer:
The reaction quotient Q for C(s) + H2O(g) ⇌ CO(g) + H2(g), given the amounts of reactants and products and the volume of the container, is calculated to be 1.12.
Explanation:
The reaction quotient, Q, expresses the relative amounts of reactants and products during the progress of a reaction at a particular moment in time. It is calculated using the same expression as the equilibrium constant, Kc, but with the concentrations (or in the case of gases, the partial pressures) that are currently present, rather than at equilibrium. For the reaction given:
C(s) + H₂O(g) ⇌ CO(g) + H₂(g)
The reaction quotient, Q, can be expressed as:
Q = ([CO][H₂]) / [H₂O]
The concentrations can be found by dividing the number of moles of each gas by the volume of the container. Since carbon is a solid, its amount does not factor into the expression for Q. In a 2.00 L container:
[H₂O] = 12.9 mol / 2.00 L = 6.45 M
[CO] = 4.00 mol / 2.00 L = 2.00 M
[H₂] = 7.20 mol / 2.00 L = 3.60 M
To calculate Q:
Q = (2.00 M × 3.60 M) / 6.45 M
Q = 7.20 M2 / 6.45 M
Q = 1.12
Thus, the reaction quotient Q for the given reaction under the specified conditions is 1.12.
Which of the following statements are true with respect to enzyme activity? Check all that apply. View Available Hint(s) Check all that apply. Enzymes speed up the reaction rate. The activation energy of a reaction increases when an enzyme is used to catalyze the reaction. Enzyme reactivity is not affected by change in pH and temperature. Enzymes are regenerated when the reaction is completed. Enzymes affect the reaction pathway by forming an enzyme-substrate complex.
Enzymes speed up reactions, are regenerated after the reaction, and form enzyme-substrate complexes. They do not increase activation energy or remain unaffected by pH and temperature changes.
Explanation:The following statements are true regarding enzyme activity:
Enzymes speed up the reaction rate. They do this by providing an alternative pathway for a reaction with a lower activation energy.Enzymes are regenerated when the reaction is completed. After an enzyme has bound to its substrate and completed its catalytic function, it's released intact and can participate in future reactions.Enzymes affect the reaction pathway by forming an enzyme-substrate complex. The active site of an enzyme accommodates the substrate molecule, leading to a temporary enzyme-substrate complex that allows the reaction to proceed more easily.However, the statement that "The activation energy of a reaction increases when an enzyme is used to catalyze the reaction" is incorrect. Enzymes lower the activation energy of a reaction rather than increasing it. Also false is "Enzyme reactivity is not affected by change in pH and temperature." Enzyme activity is very sensitive to changes in pH and temperature.
Learn more about Enzyme Activity here:https://brainly.com/question/35489510
#SPJ6
Use the bond energies provided to estimate ΔH°rxn for the reaction below. C2H4(g) + H2(g) → C2H6(g) ΔH°rxn = ? Bond Bond Energy (kJ/mol) C-C 347 C-H 414 C=C 611 C≡C 837 H-H 436 A) -128 kJ B) +98 kJ C) +700 kJ D) -102 kJ E) -166 kJ
hey there!:
C2H4
1 c=c ->611
4 C-H -> 4*414=1656
=> Ha=2267 kj
H2 : H-H 436
Hb = 436
C2H6
1 C-C 347
6 C-H 6*414=2484
=> Hc=2831
H=(Ha+Hb)-Hc=2267+436-2831 = -128kj
Answer A
Answer:
[tex]\Delta H_{rxn}^o=-128kJ[/tex]
Explanation:
Hello,
In this case, the standard enthalpy of reaction could be computed via the bond energies when both broken or made as shown below:
[tex]\Delta H_{rxn}^o=\Delta H_{broken}+\Delta H_{made}[/tex]
In this manner, we infer that at the reactants for ethene, [tex]C_2H_4[/tex] a double bond between carbons is broken as well as a bond between hydrogens (such values turn out positive). Furthermore, a single bond between carbons and two single bonds between carbon and hydrogen are made (such values turn out negative), in such a way, we develop the aforesaid equation to obtain:
[tex]\Delta H_{rxn}^o=(611kJ+436kJ)+(-347kJ-2*414kJ)\\\Delta H_{rxn}^o=-128kJ[/tex]
Best regards.
What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, C2H6? In this reaction, ethane is burned in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O). ____C2H6(g)+____O2(g)→____CO2(g)+____H2O(g) Recall that the coefficients of the final balanced equation should be whole numbers. Thus, you might need to multiply through the equation by a factor of two to obtain whole numbers in your last step. If you have trouble balancing the equation below, use the first hint to view a video of a similar equation being balanced. Then, use the rest of the hints to help you balance the equation, step-by-step. Express the coefficients as integers separated by commas. View Available Hint(s)
Answer: The coefficients are 2, 7, 4 and 6.
Explanation:
Every balanced chemical equation follows Law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.
This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.
For the given chemical reaction, the balanced equation follows:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO2(g)+6H_2O(g)[/tex]
On reactant side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
On product side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
Hence, the coefficients are 2, 7, 4 and 6.
To balance the combustion of ethane (C2H6), the correct coefficients are 2 for C2H6, 7 for O2, 4 for CO2, and 6 for H2O resulting in the equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
Explanation:The chemical equation for the combustion of ethane to form carbon dioxide and water is C₂H₆ + O₂ → CO₂ + H₂O. After balancing the number of atoms of each element on both sides of the equation, we arrive at the balanced chemical equation: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Thus, the resulting coefficients are 2, 7, 4, and 6.
Firstly, the carbon (C) atoms are balanced. As we have 2 carbon atoms on the left side, we can put the coefficient 2 in front of CO₂ on the right side.Next, we balance the hydrogen (H) atoms. Since we have 6 hydrogen atoms on the left side, we put the coefficient 3 in front of H₂O on the right side.Finally, we balance the oxygen (O) atoms. Counting the oxygen atoms on both sides, we see that we have 7 oxygen atoms on the right (4 from CO₂ and 3 from H₂O), meaning we need 7 O₂ molecules on the left side.Learn more about Chemical reaction balancing here:https://brainly.com/question/32834309
#SPJ11
Which of the following concerning electrolytes and the solubility rules is/are true? 1. The solubility rules apply only to ionic substances. 2. All soluble ionic compounds are considered strong electrolytes. 3. The solubility rules apply to aqueous and non-aqueous solutions.
Answer:
1. is true
Explanation:
The solubility rules apply only to salts, which are ionic compounds.
2. is false. A strong electrolyte is a salt that dissociates completely in solution. Not all salts dissociate completely. For example, a 0.36 mol·L⁻¹ solution dissociates as:
K₂SO₄ ⟶ K⁺ + KSO₄⁻ (30 %) + SO₄²⁻
Thus, K₂SO₄ does not dissociate completely into K⁺ and SO₄²⁻ ions.
3. is false. The solubility rules apply only to aqueous solutions.
The electrolytes and the solubility rules that are true include:
1. The solubility rules apply only to ionic substances. 2. All soluble ionic compounds are considered strong electrolytes. It should be noted that solubility rules only apply to the ability of ionic compounds to be able to dissolve in water. The solubility rule is important as it helps in the determination of the solubility of a particular substance before one attempts to create it.In a situation where the amount of solute is the same as the solubility, then it's saturated. Also, all soluble ionic compounds are considered strong electrolytes.In conclusion, the correct options are 1 and 2.
Read related link on:
https://brainly.com/question/4117609
3.5 mg of optically-enriched (-) sugar P was dissolved in 1.00 mL methanol in a 1.00 mL volumetric flask. The optical rotation was found to be (-)0.022° in a 5.0 cm cell. What is the specific rotation? (1 points) Show all your work
Specific rotation of a liquids can be determined from concentration, length of column and its optical rotation. The specific rotation of the solution in 5 cm length column and 3.5 g in 1 mL is -12.57°.
What is optical rotation?Optical rotation of a solution is the angle of rotation of a plane polarized light which passes through the solution. If the solution is in 1 decimeter column with 1g/l of concentration then it is called specific rotation.
The plane polarized light have orientation to either one direction left or right. The plane polarized light to left is leavo and that of right is called dextro.
The equation for specific rotation using density in g/ml and length of column in decimeter is as shown below:
specific rotation = [tex]\frac{\alpha }{d \times l}[/tex]
Where, l is the column length and d is density, alpha is the optical rotation.
Apply thr given values to the above equation.
5 cm = 0.5 decimeter.
specific rotation = - 0.022 /(0.5 × 3.5/1 ml)
= -12.57°.
Hence, the specific rotation of the solution in 5 cm length column and 3.5 g in 1 mL with an optical rotation of -0.022 is -12.57°.
To find more about specific rotation, refer the link below:
https://brainly.com/question/7337054
#SPJ2
(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, no new cells are formed?
Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:
[tex]5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O[/tex]
Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =[tex]\frac{0.03 g}{78 g/mol}=0.0003846 mol[/tex]
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
[tex]\frac{15}{5}\times 0.0003846 mol=0.0011538 mol[/tex] of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
Which of the following assumptions appears reasonable for the isothermal process? H20 (liq, 1 bar) → H2O (liq, 1300 bar), T = 20°C. A. AU = 0, AH 0 B. AU = 0, AH 0 C. AU #0, AH = 0 D. None of the above Hinn nf diethyl ether using the Chen's rule is
Answer : The assumptions appears reasonable for the isothermal process is, [tex]\Delta U=0[/tex] and [tex]\Delta H=0[/tex]
Explanation :
First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
The expression for internal energy is:
[tex]\Delta U=nC_vdt[/tex]
The expression for enthalpy is:
[tex]\Delta H=nC_pdt[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
n = number of moles
[tex]C_v[/tex] = specific heat capacity at constant volume
[tex]C_p[/tex] = specific heat capacity at constant pressure
[tex]dt[/tex] = change in temperature
As we know that, the term internal energy and enthalpy is the depend on the temperature and the process is isothermal that means at constant temperature.
T = constant
[tex]\dt[/tex] = 0
So, at constant temperature the internal energy and enthalpy is equal to zero. That means,
[tex]\Delta U=0[/tex] and [tex]\Delta H=0[/tex]
For an isothermal process in which the pressure changes significantly (as in the one outlined in the question, from 1 bar to 1300), the change in internal energy (ΔU) will be zero but a change in enthalpy (ΔH) will be non-zero, since the work done on the system is not constant. This corresponds to option B in the question.
Explanation:In order to understand the question, it's crucial to define two key terms: isothermal and internal energy. An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. For an ideal gas, the internal energy is a function of temperature only. Hence, in an isothermal process, the change in internal energy ΔU = 0. The enthalpy H of a system is defined as H = U + PV where U is the internal energy of the system, P is the pressure, and V is the volume.
Given that the process we're examining here involves a change in pressure, there is work done on the system (from 1 bar to 1300 bar), so the enthalpy H = U + PV will not be zero as P and V do not remain constant. Therefore, the most reasonable assumption for this process is ΔU = 0 and ΔH ≠ 0, which would correspond to option B.
Learn more about the Isothermal Process here:https://brainly.com/question/29209594
#SPJ3
Which of the following statements is true? A) If Q < K, it means the forward reaction will proceed to form more products. B) If Q > K, it means the forward reaction will proceed to form more products. C) If Q = K, it means the reaction is not at equilibrium. D) All of the above are true. E) None of the above are true.
Hey there!:
If Q = K, it means the reaction is at equilibrium.
The correct statments are as follows:
A. If Q < K, it means the forward reaction will proceed to form more products.
B. If Q > K, it means the backward reaction will proceed to form more reactants.
The statement C is true.
Hope this helps!
The correct statement is " If Q < K, it means the forward reaction will proceed to form more products."
What is the forward reaction?Any chemical change has been considered a forward reaction if the reactants reacted to produce the product on the opposite side of the arrow.
What are products?The species that result from chemical reactions have always been called products. In a chemical reaction, reactants undergo a slightly elevated transition state before becoming products.
Therefore, the correct statement is " If Q < K, it means the forward reaction will proceed to form more products."
To know more about forward reaction and products
https://brainly.com/question/8592296?referrer=searchResults
#SPJ2
Which statements are true? The oxidation number for Cu(s) is +2. An oxidizing agent gains electrons. Na+ is formed from the reduction of Na(s) . The oxidation number for Hg(????) is 0. A reducing agent gains electrons. Zn2+ is formed from the oxidation of Zn(s) .
Final answer:
The true statements are: an oxidizing agent gains electrons, a reducing agent loses electrons, and Zn2+ is formed from the oxidation of Zn(s).
Explanation:
Let's address each statement in the question about oxidation numbers and redox reactions:The oxidation number for Cu(s) is 0, not +2, because the oxidation number of a free element is always 0.An oxidizing agent gains electrons during a redox reaction.Na+ is formed from the oxidation of Na(s), not reduction, because sodium loses an electron to form Na+.The oxidation number for elemental Hg(l) is 0 since the oxidation state of a free element is 0.A reducing agent loses electrons in a redox reaction.Zn2+ is formed from the oxidation of Zn(s) since zinc loses electrons in the process.2Al(s)+Fe2O3(s)−→−heatAl2O3(s)+2Fe(l) 2Al(s)+Fe2O3(s)→heatAl2O3(s)+2Fe(l) If 26.1 kg Al26.1 kg Al reacts with an excess of Fe2O3,Fe2O3, how many kilograms of Al2O3Al2O3 will be produced?
Answer : The mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
Explanation : Given,
Mass of [tex]Al[/tex] = 26.1 Kg = 26100 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]Al_2O_3[/tex] = 32 g/mole
First we have to calculate the moles of [tex]Al[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{26100g}{26.98g/mole}=967.38moles[/tex]
Now we have to calculate the moles of [tex]Al_2O_3[/tex].
The balanced chemical reaction is,
[tex]2Al+Fe_2O_3\rightarrow Heat+Al_2O_3+2Fe[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react to give 1 mole of [tex]Al_2O_3[/tex]
So, 967.38 moles of [tex]Al[/tex] react to give [tex]\frac{967.38}{2}=483.69[/tex] moles of [tex]Al_2O_3[/tex]
Now we have to calculate the mass of [tex]Al_2O_3[/tex].
[tex]\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3[/tex]
[tex]\text{Mass of }Al_2O_3=(483.69mole)\times (101.96g/mole)=49317.0324g=49.32Kg[/tex]
Therefore, the mass of [tex]Al_2O_3[/tex] produced will be, 49.32 Kg
In the reaction of 26.1 kg of Al with an excess of Fe₂O₃, whose equation is 2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l), will be produced 49.31 kilograms of Al₂O₃.
The reaction is:
2Al(s) + Fe₂O₃(s) → heat + Al₂O₃(s) + 2Fe(l) (1)
To find the mass of Al₂O₃ produced, we need to find the number of moles of Al since it is the limiting reactant (Fe₂O₃ is in excess).
[tex] n_{Al} = \frac{m_{Al}}{A_{Al}} [/tex] (2)
Where:
[tex]m_{Al}[/tex]: is the mass of Al = 26.1 kg = 26100 g
[tex]A_{Al}[/tex]: is the atomic mass of Al = 26.982 g/mol
The number of moles of Al is (eq 2):
[tex]n_{Al} = \frac{m_{Al}}{A_{Al}} = \frac{26100 g}{26.982 g/mol} = 967.31 \:moles[/tex]
From equation (1) we have that 2 moles of Al react with 1 mol of Fe₂O₃ to form 1 mol of Al₂O₃(s), so the number of moles of Al₂O₃ produced is:
[tex]n_{Al_{2}O_{3}} = \frac{1 \:mol \:Al_{2}O_{3}}{2 \:moles \:Al}*967.31 \:moles \:Al = 483.66 \:moles[/tex]
Finally, the mass of Al₂O₃ in kilograms is:
[tex]m = n_{Al_{2}O_{3}}*MM = 483.66 \:moles*101.96 \:g/mol = 49313.5 g = 49.31 kg[/tex]
Therefore, will be produced 49.31 kilograms of Al₂O₃.
Read more about limiting reactants here:
https://brainly.com/question/14225536?referrer=searchResultshttps://brainly.com/question/7144022?referrer=searchResultsI hope it helps you!
Ethanol has a Kb of 1.22 Degrees C/m and usually boils at 78.4 Degrees Celcius. How many mol of an nonionizing solute would need to be added to 47.84 g ethanol in order to raise the boiling point to 86.30?
Answer:
0.3097 moles of an nonionizing solute would need to be added.
Explanation:
Molal elevation constant = [tex]k_b=1.22^oC/m[/tex]
Normal boiling point of ethanol = [tex]T_o=78.4^oC[/tex]
Boiling of solution =[tex]T_b=86.30^oC[/tex]
Moles of nonionizing solute = n
Mass of ethanol (solvent) = 47.84 g
Elevation boiling point:
[tex]\Delta T_b=T_b-T_o[/tex]
[tex]\Delta T_b=86.30^oC-78.4^oC=7.9^oC[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]m=\frac{\text{Moloes of solute}}{\text{Mass of solvent(kg)}}[/tex]
[tex]7.9^oC=1.22^oC/m\times \frac{n}{0.04784 kg}[/tex]
n = 0.3097 mol
0.3097 moles of an nonionizing solute would need to be added.
Final answer:
To raise the boiling point of ethanol to 86.30°C, 0.310 moles of a nonionizing solute need to be added to 47.84 g of ethanol, using the boiling point elevation formula and the given Kb of ethanol.
Explanation:
To calculate the number of moles of a nonionizing solute needed to raise the boiling point of ethanol to 86.30°C, we use the boiling point elevation formula: ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the ebullioscopic constant of ethanol, and m is the molality of the solution.
First, determine the change in boiling point (ΔT): ΔT = final boiling point - initial boiling point = 86.30°C - 78.4°C = 7.9°C.
Use the given Kb for ethanol, 1.22°C/m, and solve for molality (m): m = ΔT / Kb = 7.9°C / 1.22°C/m = 6.48 m.
To find the number of moles of solute, relate molality to the mass of solvent in kilograms: molality (m) = moles of solute/kg of solvent. Therefore, moles of solute = m × kg of solvent. The mass of ethanol is 47.84 g or 0.04784 kg, thus moles of solute = 6.48 × 0.04784 kg = 0.310 moles.
Ammonium chloride, NH4Cl, is a salt formed from the neutralization of the weak base ammonia, NH3, with the strong acid hydrochloric acid. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.176 M solution of ammonium chloride at 25∘C?
Answer:
The pH of a 0.176 M solution of ammonium chloride is 4.9902.
Explanation:
Given:
Kb for ammonia = 1.8×10⁻⁵
Since ammonium ions are the conjugate acid of ammonia.
Thus, Ka for ammonium ions;
Ka = Kw/Kb = 10⁻¹⁴ / 1.8×10⁻⁵ = 5.5556×10⁻¹⁰
Given concentration of Ammonium chloride (C) = 0.176 M
Thus, for weak acids,
[tex]\left[H^+ \right]=\sqrt{K_a\times C}[/tex]
[tex]\left[H^+ \right]=\sqrt{5.5556\times 10^{-10}\times 0.176}[/tex]
[tex]\left[H^+ \right]=0.9777\times 10^{-5}[/tex]
pH is:
[tex]pH\ of\ the\ solution=-log\left[H^+ \right][/tex]
[tex]pH\ of\ the\ solution=-log\left(0.9777\times 10^{-5} \right)[/tex]
pH = 4.9902
Thus,
The pH of a 0.176 M solution of ammonium chloride is 4.9902.
The pH of a 0.176 M solution of ammonium chloride at 25 degrees Celsius is 11.01, calculated by using the equilibrium equation of the hydrolysis of the ammonium ion and taking into account the neutrality of the reaction.
Explanation:The question requires calculating the pH of a 0.176 M solution of ammonium chloride (NH4Cl), which is formed from the neutralization of the weak base ammonia (NH3) by the strong acid hydrochloric acid (HCl).
To find the pH, it is necessary to find the concentration of the hydronium ion [H3O+]} in the solution. The neutrality of the reaction means that when NH3 reacts with HCl, it forms its conjugate acid NH4+, which hydrolyzed in water. The equilibrium equation for the hydrolysis NH4+ + H20 <-> NH3 + H3O+ and knowing the value of Kb for ammonia (1.8×10−5), you can calculate Ka for the ammonium ion, which helps calculate [H3O+]. Using the expression Kw = Ka x Kb, since Kw is 1.0 x 10^-14 at 25 degrees Celcius, you can find Ka = Kw / Kb, which equals 5.56 x 10^-10. Using the hydrolysis reaction equilibrium equation, Ka = [NH3][H3O+] / [NH4+], and knowing [NH4+] = 0.176M (from the problem statement), you isolate [H3O+] and find it equals 9.79 x 10^-12. Finally, using the pH=-log10[H3O+], calculate the pH to be 11.01.
Learn more about pH calculation here:https://brainly.com/question/34432529
#SPJ6
Coal can be used to generate hydrogen gas (a potential fuel) by thefollowing endothermic reaction.
C(s) + H2O (g) <==> CO(g) + H2(g)
if this reaction mixture is at equilibrium, predict whether each ofthe following will result in the formation of additional hydrogengas, the formation of less hydrogen gas, or have no effect on thequantity og hydrogen gas.
a. adding more C to the reaction mixture
b. adding more H2O to the reaction mixture
c. raising the temperature of the reaction mixture
d. increasing the volume of the reaction mixture
e. adding a catalyst to the reaction mixture
f. adding an inert gas to the reaction mixture
Answer:The statements b,c,d would lead to increased formation of more hydrogen gas and statements a, e and f would lead to unchanged hydrogen concentration
Explanation:
Lechateliers principle can be used here to determine the effect of changes observed in the system.
Lechateliers principle states that if any reaction at equilibrium is subjected to change in concentration, temperature and pressure or even in reaction conditions then the equilibrium of the reaction would shift in such a way so that it can oppose the change .
So if any disturbance is caused to a reaction at equilibrium hence the equilibrium of reaction would shift in such a way so that it can counter balance the change caused to the reaction.
The above reaction is following:
C(s)+H₂O(g)→CO(g)+H₂(g)
The enthalpy change of this reaction is positive and hence the reaction is endothermic in nature.
So the given changes would lead to the following results:
a The addition of more amount of carbon C(s) would not lead to any further formation of hydrogen because carbon is added in solid state and Hydrogen gas is in gaseous state so the equilibrium for this given reaction would only change on addition of gaseous reactants as that would only lead to change in concentration.
b Since H₂O(g) is in gaseous state and a reactant and hence the addition of H₂O(g) that is more reactant would lead to more formation of hydrogen gas according to lechatelier principle. The equilibrium would shift in such a way so that it can decrease the concentration of added H₂O(g) hence it would form H₂(g).
c Since the above reaction is endothermic in nature hence increasing the temperature of reaction would also shift the equilibrium of reaction towards more formation of H₂(g) that is in forward direction.
d When we increase the volume of reaction mixture that is we are increasing the amount of reactants hence the reaction would shift towards more formation of hydorgen gas.
e The catalyst does not change the position of equilibrium and hence no shift in position of equilibrium would be observed.So amount of hydrogen gas formed would remain unchanged.
f The addition of inert gas would not lead to any change to the reaction and equilibrium would be unaffected. Hence the formation of hydrogen gas would remain unchanged.
The study of chemicals and the bond is called chemistry. When the amount of the reactant and the product get equal is said to be equilibrium.
The correct answer is b, c, d would lead to increased formation of more hydrogen gas, and statements a, e and f would lead to unchanged hydrogen concentration.
According to the principle can be used here to determine the effect of changes observed in the system.
What is an endothermic reaction?Those reactions that require energy or temperature to complete the reaction are called endothermic reactions.The principle states that if any reaction at equilibrium is subjected to change in Concentration, Temperature, Pressure, or even in reaction conditions then the equilibrium of the reaction would shift in such a way so that it can oppose the change.So if any disturbance is caused to a reaction at equilibrium hence the equilibrium of reaction would shift in such a way so that it can counterbalance the change caused to the reaction.
The reaction is as follows:-
[tex]C(s)+H_2O(g)---->CO(g)+H_2(g)[/tex]
The enthalpy change of this reaction is positive and hence the reaction is endothermic in nature.
So the given changes would lead to the following results:
A. The addition of more amount of carbon C(s) would not lead to any further formation of hydrogen because carbon is added in solid-state. Hydrogen gas is in the gaseous state so the equilibrium for this given reaction would only change on the addition of gaseous reactants as that would only lead to a change in concentration.
B. Since H₂O(g) is in a gaseous state and a reactant and hence the addition of H₂O(g) that is more reactant would lead to more formation of hydrogen gas according to the principle. The equilibrium would shift in such a way so that it can decrease the concentration of added H₂O(g) hence it would form H₂(g).
C . Since the above reaction is endothermic in nature hence increasing the temperature of the reaction would also shift the equilibrium of reaction towards more formation of H₂(g) that is in the forward direction.
D. When we increase the volume of the reaction mixture that is we are increasing the number of reactants hence the reaction would shift towards more formation of hydrogen gas.
E. The catalyst does not change the position of equilibrium and hence no shift in the position of equilibrium would be observed. So the amount of hydrogen gas formed would remain unchanged.
F. The addition of inert gas would not lead to any change to the reaction and equilibrium would be unaffected. Hence the formation of hydrogen gas would remain unchanged.
Hence, the correct answer is mentioned above.
For more information about the endothermic reaction, refer to the link:-https://brainly.com/question/2192784
4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/mol, D 0.79 g/mL), 25 mL of concentrated sulfuric acid (MW 98.08 g/mol, D 1.84 g/mL), and 5.6 g of 1,4-dimethoxybenzene (MW 138.17 g/mol) together. Calculate the percent yield of this reaction.
Answer:
Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.
Explanation:
Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.
Mass = Density × Volume
Mass of t-butyl alcohol = [tex]0.79 g/mL\times 10.4 mL=8.219 g[/tex]
Moles of t-butyl alcohol =[tex]\frac{8.219 g}{74.12 g/mol}=0.11084 mol[/tex]
Moles of 1,4-dimethoxybenzene = [tex]\frac{5.6 g}{138.17 g/mol}=0.04052 mol[/tex]
According to reaction 2 mol of t-butyl alcohol reacts with 1 mol of 1,4-dimethoxybenzene.
Then 0.11084 moles of t-butyl alcohol will react with :
[tex]\frac{1}{2}\times 0.11084 mole=0.05542 mol[/tex] of 1,4-dimethoxybenzene.
This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.
According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of 1,4-di-t-butyl-2,5-dimethoxybenzene.
Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:
[tex]\frac{1}{1}\times 0.04052 mol= 0.04052 mol[/tex] of 1,4-di-t-butyl-2,5-dimethoxybenzene.
Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:
0.04052 mol × 250.37 g/mol = 10.144 g
Percentage yield:
[tex]\frac{Experimental}{Theoretical}\times 100[/tex]
Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:
Experimental yield = 4.2 g
Theoretical yield = 10.144 g
[tex]\frac{4.2 g}{10.144 g}\times 100=41.40\%[/tex]
The percent yield of the reaction is approximately 41.52%.
The correct format for the answer is as follows:
First, we need to calculate the moles of the limiting reactant to determine the theoretical yield. The balanced chemical equation for the synthesis of 1,4-di-t-butyl-2,5-dimethoxybenzene is not provided, but we can assume that one mole of 1,4-dimethoxybenzene reacts with two moles of t-butyl alcohol to produce one mole of the product.
Let's calculate the moles of each reactant:
For t-butyl alcohol:
[tex]\[ \text{Moles of t-butyl alcohol} = \frac{\text{Volume (mL)} \times \text{Density (g/mL)}}{\text{Molecular Weight (g/mol)}} \] \[ \text{Moles of t-butyl alcohol} = \frac{10.4 \text{ mL} \times 0.79 \text{ g/mL}}{74.12 \text{ g/mol}} \] \[ \text{Moles of t-butyl alcohol} = \frac{8.216 \text{ g}}{74.12 \text{ g/mol}} \] \[ \text{Moles of t-butyl alcohol} \approx 0.1108 \text{ mol} \][/tex]
For 1,4-dimethoxybenzene:
[tex]\[ \text{Moles of 1,4-dimethoxybenzene} = \frac{\text{Mass (g)}}{\text{Molecular Weight (g/mol)}} \] \[ \text{Moles of 1,4-dimethoxybenzene} = \frac{5.6 \text{ g}}{138.17 \text{ g/mol}} \] \[ \text{Moles of 1,4-dimethoxybenzene} \approx 0.0405 \text{ mol} \][/tex]
Since the reaction requires two moles of t-butyl alcohol for every mole of 1,4-dimethoxybenzene, t-butyl alcohol is in excess, and 1,4-dimethoxybenzene is the limiting reactant.
Now, we calculate the theoretical yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:
[tex]\[ \text{Theoretical yield (g)} = \text{Moles of limiting reactant} \times \text{Molecular Weight of product} \] \[ \text{Theoretical yield (g)} = 0.0405 \text{ mol} \times 250.37 \text{ g/mol} \] \[ \text{Theoretical yield (g)} \approx 10.11955 \text{ g} \]The actual yield is given as 4.2 g.[/tex]
Finally, we calculate the percent yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield (g)}}{\text{Theoretical Yield (g)}} \right) \times 100\% \] \[ \text{Percent Yield} = \left( \frac{4.2 \text{ g}}{10.11955 \text{ g}} \right) \times 100\% \] \[ \text{Percent Yield} \approx 41.52\% \][/tex]