Answer: A. PH=4.50
Explanation:
As we know that pH=-log(H3O+)
pH=-log(3.2x10^-5)
PH= -log(3.2)+log(10^-5)
=-(0.50+(-5))
PH= -(0.5-5)=-(-4.5)
pH=4.50
B. If we know pH , then [H3O+]= 10^-pH
To find the pH of a substance, use pH = -log[H3O+]. Given [H3O+] = 3.2 × 10-5, the pH ≈ 4.49. To find hydrogen ion concentration from pH, use [H3O+] = 10^-pH.
pH is a measure of the acidity or alkalinity of a solution, defined as the negative logarithm of the hydrogen ion concentration (expressed in molarity) of a solution.
To find the pH of a substance, you can use the formula: pH = -log[H3O+].
a. Given [H3O+] = 3.2 × 10-5, the pH is calculated as follows:
pH = -log(3.2 × 10-5)pH ≈ 4.49b. To find the hydrogen ion concentration given a pH value, you rearrange the formula as: [H3O+] = 10-pH.
For example, if the pH of a solution is 3:
[H3O+] = 10-3[H3O+] = 0.001 MTherefore, solutions with a pH of less than seven are acidic, while those with a pH greater than seven are basic (alkaline). The pH scale ranges from 0 to 14 and a pH of 7 is considered neutral.
The catalytic converter, a required component for automobile exhaust emission control systems, converts carbon monoxide into A. Pure carbon and oxygen B. Carbon dioxide C. Carbonic acid D. Carbon monoxide (the catalytic converter is designed to remove gases other than carbon monoxide)
Answer:
B. Carbon dioxide
Explanation:
A catalytic converter is designed and placed in the exhaust of most automobiles to remove harmful gases from escaping into the environment.
The converter uses palladium and platinum which are both catalyst to reduce harmful gases such as nitrogen oxide which causes acid rain, carbon monoxide which affects human haemoglobin and other hydrocarbons into less harmful ones.
The catalytic converter helps to convert carbon monoxide, a product of incomplete combustion of hydrocarbons into less harmful carbondioxde. Carbon monoxide is a very toxic gas. It causes harm to both plants and animals as well.
An achiral hydrocarbon A of molecular formula C7H12 reacts with two equivalents of H2 in the presence of Pd-C to form CH3CH2CH2CH2CH(CH3)2. One oxidative cleavage product formed by the treatment of A with O3 is CH3COOH. Reaction of A with H2 and Lindlar catalyst forms B, and reaction of A with Na, NH3 forms C. Identify compounds A, B, and C. Be sure to answer all parts.
Answer:
A) 5-methylhex-2-yne
B) (2Z)-5-methylhex-2-ene
C) (2E)-5-methylhex-2-ene
Explanation:
The given compound must be alkyne as it is undergoing reduction with two equivalents of hydrogen molecule.
Also as it is giving acetic acid on oxidative ozonolysis, it must have triple bond after two carbons in the chain.
The structure of hydrocarbon formed after reduction will give us the structure of alkyne by these information.
Reaction with hydrogen molecule in presence of Lindlar's catalyst gives cis alkene.
Reaction with hydrogen molecule in presence of Na, ammonia gives trans alkene.
The structure of compound is shown in the figure
The achiral hydrocarbon (A) with a molecular formula C7H12 is identified as Hept-1-yne. Upon reacting with a Lindlar catalyst and H2, it forms Hept-1-ene (B). When Hept-1-yne is treated with Sodium in ammonia (Na, NH3), it forms 1-heptyne (C).
Explanation:The hydrocarbon A that has a molecular formula of C7H12 should be Hept-1-yne in light of the fact that Hept-1-yne upon hydrogenation, utilizing Pd-C as a catalyst and two equivalents of H2, forms 4-methyl hexane, which is exact to what was stated in the question. When Hept-1-yne is treated with ozone (O3), it gives two oxidative cleavage products, one being CH3COOH (acetic acid). The reaction gives us a clue about the presence of a triple bond at the end of the heptane chain. Hence, the structure of compound A (Hept-1-yne) is identified as CH3-(CH2)4-C≡CH.
Compound B can be identified as Hept-1-ene. This is due to the fact that Hept-1-yne, upon reacting with a Lindlar catalyst and H2, forms Hept-1-ene. This conversion is a result of the partial reduction of the triple bond to a double bond.Lastly, compound C can be identified as 1-heptyne. This is because when we treat Hept-1-yne with Sodium in ammonia (Na, NH3, it selectively reduces the triple bond to a trans double bond, a process known as dissolving metal reduction.
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An initially evacuated 1.5 m tank is fed (adiabatically) with steam from a line available at a constant 15 MPa and 400 °C until the tank pressure reaches 15 MPa. What is the final mass of water in the tank in kg?
Answer:
95.8 kg
Explanation:
At the end of the feeding process, there is steam in the tank at 15 MPa and 400ºC because the process is adiabatic. So, use the steam tables (In this case I use data from van Wylen Six Edition, table B.13) in order to get the specific volume of superheated steam.
The specific volume data reported is [tex]v=0.01565\frac{m^{3}}{kg}[/tex]
The mass can be calculated with the definition of specific volume:
[tex]v=\frac{V}{m}\\m=\frac{V}{v}=\frac{1.5m^{3}}{0.01565\frac{m^{3}}{kg}} =95.8kg[/tex]
The limiting reactant is completely consumed in a chemical reaction. (T/F)
Answer: Yes
Explanation:
Limiting reagent is the reagent which limits the formation of product as it gets completely consumed in the reaction.
Excess reagent is the reagent which is left unreacted in the reaction.
For example: [tex]2HCl+Ca\rightarrow CaCl_2+H_2[/tex]
If there are 2 moles of [tex]HCl[/tex] and 2 moles of [tex]Ca[/tex]
As can be seen from the chemical equation,
2 moles of hydrochloric acid react with 1 mole of calcium.
Thus 2 moles of [tex]HCl[/tex] will completely react with 1 mole of calcium and (2-1)=1 mole of calcium will remain as such.
Thus HCl is the limiting reagent as it limits the formation of product and calcium is the excess reagent as it is left unreacted.
Final answer:
The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction. It is completely consumed in the reaction.
Explanation:
The limiting reactant (or limiting reagent) is the reactant that determines the amount of product that can be formed in a chemical reaction. The reaction proceeds until the limiting reactant is completely used up. The other reactant or reactants are considered to be in excess. To determine the limiting reactant, you need to compare the amount of each reactant present in the reaction to the stoichiometric ratios in the balanced chemical equation. The reactant with the smallest amount is the limiting reactant.
In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL + 10.00 mL of the acid have been added?
Answer: The pH of the solution is 1.136
Explanation:
To calculate the moles from molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
For ammonia:Molarity of ammonia = 0.3764 M
Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol[/tex]
For nitric acid:Molarity of nitric acid = 0.3838 M
Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L
Putting values in above equation, we get:
[tex]0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol[/tex]
After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol
For the reaction of ammonia with nitric acid, the equation follows:
[tex]NH_3+HNO_3\rightarrow NH_4NO_3[/tex]
At [tex]t=0[/tex] 0.0178 0.022
Completion 0 0.0042 0.0178
As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.
To calculate the pH of the reaction, we use the equation:
[tex]pH=-\log[H^+][/tex]
where,
[tex][H^+]=\frac{0.0042mol}{0.05741L}=0.0731M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(0.0731)\\\\pH=1.136[/tex]
Hence, the pH of the solution is 1.136
1. In an equilibrium experiment, acetic acid (which is a weak acid) is mixed with sodium acetate (a soluble salt), with methyl orange as an indicator. Explain this phenomenon by using the common ion effect. Include equations in your explanation.
Explanation:
It is known that acetic acid is a weak acid. It's equilibrium of dissociation will be represented as follows.
[tex]CH_{3}COOH(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)[/tex]
On the other hand, sodium acetate ([tex]CH_{3}COONa[/tex]) is a salt of weak acid, that is, [tex]CH_{3}COOH[/tex] and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.
Since, acetic acid is a weak acid but still it is an acid. So, when methyl orange is added in a solution of acetic acid then it given a reddish-orange color because of its acidity.
When sodium acetate is mixed into this solution then it will dissociate as follows.
[tex]CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)[/tex]
As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.
So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.
As a result, there will be decrease in the concentration of hydronium ions. As a result, there will be increase in the pH of the system.
Hence, color of methyl orange will change from reddish orange to yellow. This shift in equilibrium is due to the common ion which is [tex]CH_{3}COO^{-}[/tex] ion.
A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from the nitrogen atoms and compound Y is produced from the metal atoms in the reactant. What are the compounds X and Y? X is nitrogen dioxide, and Y is a metal halide. X is nitrogen dioxide, and Y is a metal oxide. X is nitrogen gas, and Y is a metal sulfate. X is nitrogen gas, and Y is a metal oxide.
Answer:
The correct answer is: X is nitrogen dioxide, and Y is a metal oxide
Explanation:
Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.
The general equation is given as:
[tex]4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x[/tex]
Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.
In a combustion reaction, a compound containing nitrogen and a metal typically forms nitrogen dioxide (compound X) from the nitrogen atoms and a metal oxide (compound Y) from the metal atoms.
Explanation:In the context of your question about how a compound containing nitrogen and a metal reacts in a combustion reaction, the outcome depends on the specific reactant. Usually, compounds containing nitrogen atoms tend to form nitrogen oxides under high heat or combustion conditions, with nitrogen dioxide (NO2) being a common example. This would be compound X.
Regarding the metal component, in a combustion reaction, metals commonly react with oxygen in the environment to produce metal oxides. This would make compound Y a metal oxide. Therefore, the correct pair of products according to your question tends to be: X is nitrogen dioxide, and Y is a metal oxide.
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Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X2O3 and write a balanced equation for the reaction.
Answer:The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Explanation:
Let the molecular formula of the oxide of metal be [tex]X_2O_y[/tex]
[tex]X_2O_y+yCO\rightarrrow yCO_2+2X[/tex]
Mass of metal product = 1.68 g
Moles of metal X =[tex]\frac{1.68 g}{55.9 g/mol}=0.03005 mol[/tex]
1 mol of metal oxide produces 2 moles of metal X.
Then 0.03005 moles of metal X will be produced by:
[tex]\frac{1}{2}\times 0.03005 mol=0.01502 mol[/tex] of metal oxide
Mass of 0.01502 mol of metal oxide = 2.40 g (given)
[tex]0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g[/tex]
y = 2.999 ≈ 3
The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:
[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]
Final answer:
To show the simplest formula of the oxide is X2O3, we calculate the moles of metal (X) and oxygen from given masses, find their ratio, and deduce the empirical formula. The balanced equation for the reaction with carbon monoxide is X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g).
Explanation:
To prove that the simplest formula of the oxide is X2O3, first we need to calculate the moles of metal X produced. Since the molar mass of X is given as 55.9 g/mol, we divide the mass of metal product (1.68 g) by the molar mass of X to obtain the number of moles:
moles of X = 1.68 g / 55.9 g/mol = 0.03005 mol
We know that the initial mass of the oxide is 2.40 g and the product (X) is 1.68 g, so the mass of oxygen in the oxide is:
mass of O = 2.40 g - 1.68 g = 0.72 g
rationalizing the ratio, we get approximately 2:3
Thus, the empirical formula of the oxide is X2O3.
Balanced Equation for the Reaction
The balanced equation for the reaction of metal X's oxide with carbon monoxide to obtain metal X and carbon dioxide is:
X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g)
This equation shows that the oxide of metal X reacts with carbon monoxide in a 1:3 mole ratio to produce the pure metal and carbon dioxide in a 2:3 mole ratio.
True or False Titanium's corrosion resistance is so strong that even titanium with oxygen impurities does no reduction in corrosion resistance.
Answer :
true
Explanation:
when titanium react with oxygen it form TiO₂ Titanium oxide which is passive in nature it forms layer of TiO₂ when large amount of oxygen is passed through titanium at very high temperature it does not react with oxygen impurities for this reason titanium mostly used in aerospace and chemical industries
Calculate the pH of a 0.22 M ethylamine solution.
Answer:
answer is 12.18
Explanation:
(C2H5NH2, Kb = 5.6 x 10-4.)
Answer:
pH = 10.1
Explanation:
For weak base solutions [OH] = SqrRt([Base]·Kb
Then, from pH + pOH = 14 => pH = 14 - pOH
[OH} = SqrRt[(0.22)(5.6 x 10⁻⁴)] = 3.91
pH = 14 - 3.91 = 10.1
Part C Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm. What is the radius of a gallium atom? Express your answer numerically in picometers. View Available Hint(s) radius = nothing pm p m Submit
Final answer:
The radius of a gallium atom in a primitive cubic unit cell, with an edge length of 362 pm, is calculated to be 181 picometers by dividing the edge length by 2.
Explanation:
The question asks for the radius of a gallium atom given that gallium crystallizes in a primitive cubic unit cell with an edge length of 362 pm. In a primitive cubic unit cell, the atoms are located at the corners of the cube, and the length of the edge of the cube is equal to twice the atomic radius. Therefore, to find the radius of the gallium atom, we divide the edge length by 2.
Radius of gallium atom = edge length / 2 = 362 pm / 2 = 181 pm.
This calculation reveals that the radius of a gallium atom is 181 picometers in a primitive cubic unit cell structure.
vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compare this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?
Answer:
q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.
Explanation:
1. Heat absorbed
q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ
2. Change in volume
V(water) = 0.018 L
pV = nRT
1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K
V = 30.62 L
ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L
3. Work done
w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm
w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ
4. Why the difference?
Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.
The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59 2 C What is the mass of the water? Express your answer to two significant figures
Answer : The mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]
Explanation :
In this case the 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_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of iron metal = [tex]0.45J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of iron metal = 32.3 g
[tex]m_2[/tex] = mass of water = ?
[tex]T_f[/tex] = final temperature of mixture = [tex]59.2^oC[/tex]
[tex]T_1[/tex] = initial temperature of iron metal = [tex]21.9^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]63.5^oC[/tex]
Now put all the given values in the above formula, we get
[tex]32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC[/tex]
[tex]m_2=30.16g\approx 3.0\times 10^1g[/tex]
Therefore, the mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]
The mass of the water in the insulated container is 30.04 g
Data obtained from the questionMass of iron (Mᵢ) = 32.2 gTemperature of iron (Tᵢ) = 21.9 °CTemperature of water (Tᵥᵥ) = 63.5 °C Equilibrium temperature (Tₑ) = 59.2 °C Specific heat capacity of iron (Cᵢ) = 0.45 J/gºCSpecific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC Mass of water (Mᵥᵥ) =? How to determine the mass of waterHeat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = MᵢC(Tₑ – Mᵢ)
Mᵥᵥ × 4.184 (63.5 – 59.2) = 32.2 × 0.45(59.2 – 21.9)
Mᵥᵥ × 4.184(4.3) = 14.49(37.3)
Clear bracket
Mᵥᵥ × 17.9912 = 540.477
Divide both side by 17.9912
Mᵥᵥ = 540.477 / 17.9912
Mᵥᵥ = 30.04 g
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Calculation of Original pH from Final pH after Titration A biochemist has 100 mL of a 0.10 M solution of a weak acid with a pKa of 6.3. She adds 6.0 mL of 1.0 M HCl, which changes the pH to 5.7. What was the pH of the original solution?
Answer:
6.9
Explanation:
A weak acid dissociates in an equilibrium reaction, thus, it is in equilibrium with its conjugate base:
HA ⇄ H⁺ + A⁻
The equilibrium constant (Ka) can be calculated, where Ka = [H⁺]*[A⁻]/[HA]. Using the -log form, we can also have pKa = -logKa. By the Handerson-Halsebach (HH) equation, the relation between pH and pKa is:
pH = pKa + log[A⁻]/[HA]
So, when pH = 5.7, for this acid, the ratio of [acid]/[base] ([HA]/[A-]) is:
5.7 = 6.3 + log[A⁻]/[HA]
log[A⁻]/[HA] = -0.6
log[HA]/[A⁻] = 0.6
[HA]/[A⁻] = [tex]10^{0.6}[/tex]
[HA]/[A⁻] = 3.98 = 4.0
If the ratio of acid and base is 4 to 1, it means that 80%(4/5) of the acid is protonated after the addition of the HCl.
The initial number of moles of the weak acid was: 100 mL * 0.10 M = 10 mmol, so after the addition of HCl, 8 mmol is in the acid form (80% of 10). It was added 6.0 mmol of HCl (6.0 mL*1.0M). Thus, 6.0 mmol of H+ was added and reacted with the conjugate base of the weak acid.
For the mass conservation, the initial amount of the protonated weak acid must be 2.0 mmol (8 - 6), and the number of moles of the conjugate base was 8.0 mmol. Using the HH equation:
pH = 6.3 + log(8/2)
pH = 6.3 + 0.6
pH = 6.9
The original pH of the weak acid solution was 6.3 which can be determined using the pKa value and the Henderson-Hasselbalch equation. The addition of HCl does not affect this original pH.
Explanation:A weak acid's initial pH can be calculated using the pKa of the acid and its initial concentration (0.10 M in this case) through the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). However, a weak acid exists in equilibrium with its conjugate base and there is no added base here initially, so the pH equals the pKa, making the original pH 6.3. The addition of HCl doesn't matter in this context because the question asks for the original pH, not the final pH.
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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
CaCO3 + 2HCl ⟶CaCl2 + H2O + CO2
A) How many grams of calcium chloride will be produced when 26.0g of calcium carbonate are combined whith 12.0g of hydrochloric acid?
B) Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?
Answer: a) 18.3 grams
b) [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
Explanation:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of calcium carbonate}=\frac{26g}{100g/mol}=0.26moles[/tex]
[tex]\text{Number of moles of hydrochloric acid}=\frac{12g}{36.5g/mol}=0.33moles[/tex]
According to stoichiometry:
2 mole of [tex]HCl[/tex] react with 1 mole of [tex]CaCO_3[/tex]
0.33 moles of [tex]HCl[/tex] will react with=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCO_3[/tex]
Thus [tex]HCl[/tex] is the limiting reagent as it limits the formation of product and [tex]CaCO_3[/tex] is the excess reagent.
2 moles of [tex]HCl[/tex] produce = 1 mole of [tex]CaCl_2[/tex]
0.33 moles of [tex]HCl[/tex] produce=[tex]\frac{1}{2}\times 0.33=0.165moles[/tex] of [tex]CaCl_2[/tex]
Mass of [tex]CaCl_2=moles\times {\text{Molar Mass}}=0.165\times 111=18.3g[/tex]
As 0.165 moles of [tex]CaCO_3[/tex] are used and (0.33-0.165)=0.165 moles of [tex]CaCO_3[/tex] are left unused.
Mass of [tex]CaCO_3[/tex] left unreacted =[tex]moles\times {\text {Molar mass}}=0.165\times 100=16.5g[/tex]
Thus 18.3 g of [tex]CaCl_2[/tex] are produced. [tex]CaCO_3[/tex] is the excess reagent and 16.5g of [tex]CaCO_3[/tex] will remain after the reaction is complete.
Be sure to answer all parts. Carry out the following operations as if it were a calculation of real experimental results. Express the answer with the correct number of significant figures. (3.26 × 10−3 mg) − (7.88 × 10−5 mg ) Answer Units mg Enter your answer in standard form. Do not use scientific notation.
Answer : The answer in standard form is, 0.00318 mg.
Explanation :
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the addition and subtraction is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
As we are given :
[tex](3.26\times10^{-3})mg-(7.88\times 10^{-5})mg[/tex]
First we have to convert scientific notation into standard form.
[tex]\Rightarrow 0.00326mg-0.0000788mg[/tex]
[tex]\Rightarrow 0.00318mg[/tex]
As per rule, the least precise number present after the decimal point is 5. So, the answer will be, 0.00318 mg.
Consider the following reactions: A: Uranium-238 emits an alpha particle B: Plutonium- 239 emits an alpha particle C: thorium-239 emits a beta particle
a. Rank the resulting nucleus by atomic number, from highest to lowest
b. Rank the resulting nucleus by the number of neutrons, from most to least
Answer:
Rank the resulting by neutrons, from most to least
The boiling points of some group 7A hydrides are tabulated below.
gas b.p. (°C)
NH3 –33
PH3 –88
AsH3 –62
Which intermolecular force or bond is responsible for the high boiling point of NH3 relative to PH3 and AsH3?A) Dipole/induced dipole force B) Covalent bonding C) Dipole-dipole force D) Induced dipole/induced dipole force E) Hydrogen bonding
hey there!:
In the case of ammonia, the hydrogen bond is formed using the lone pair present in nitrogen and the hydrogen having δ+ charge (due to bonding with electronegative N) of another ammonia molecule. Thus the inter molecular attraction increases which in turn increases the boiling point .
Option B is the correct answer.
Hope this helps!
The intermolecular force or bond which is responsible for the high boiling point of NH₃ relative to PH₃ and AsH₃ is covalent bonding.
What is covalent bonding?
Covalent bonding is defined as a type of bonding which is formed by the mutual sharing of electrons to form electron pairs between the two atoms.These electron pairs are called as bonding pairs or shared pair of electrons.
Due to the sharing of valence electrons , the atoms are able to achieve a stable electronic configuration . Covalent bonding involves many types of interactions like σ bonding,π bonding ,metal-to-metal bonding ,etc.
Sigma bonds are the strongest covalent bonds while the pi bonds are weaker covalent bonds .Covalent bonds are affected by electronegativities of the atoms present in the molecules.Compounds having covalent bonds have lower melting points as compared to those with ionic bonds.
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If the tip of the syringe, "The Titrator", was not filled with NaOH before the initial volume reading was recorded, would the concentration of acetic acid in vinegar of that trial be greater than or less than the actual concentration? Please explain your answer.
Answer:
The concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
Explanation:
"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.
Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.
So, in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.
So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.
If the syringe tip was not filled with NaOH before recording the initial volume, the concentration of acetic acid calculated would be less than the actual concentration.
If the tip of the syringe was not filled with NaOH before the initial volume reading was recorded, the concentration of acetic acid in vinegar calculated from that trial would be less than the actual concentration. This is because the actual volume of NaOH dispensed during titration would be over-reported.
When the titration is performed, the volume of NaOH required to reach the equivalence point appears larger than it truly is, leading to a miscalculation of the moles of NaOH used. Consequently, this will result in the calculated concentration of acetic acid being lower than its true value.
In titration, ensuring that the titrant (in this case, NaOH) is ready and properly measured is crucial for accurate results. The initial volume reading must be correct to avoid errors in determining the volume of NaOH added, which directly affects the accuracy of the acetic acid concentration estimation.
The equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K. CH4 (g) + CCl4 (g) 2 CH2Cl2 (g) Calculate the equilibrium concentrations of reactants and product when 0.377 moles of CH4 and 0.377 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.
Final answer:
To calculate the equilibrium concentrations of reactants and product, we need to use the given equilibrium constant, Kc, and the initial concentrations of the reactants.
Explanation:
To calculate the equilibrium concentrations of reactants and product, we need to use the given equilibrium constant, Kc, and the initial concentrations of the reactants. The balanced equation for the reaction is CH4 (g) + CCl4 (g) ⇌ 2 CH2Cl2 (g).
First, calculate the initial moles of each reactant by multiplying the initial concentration by the volume of the vessel. Then, we can set up an ICE (Initial-Change-Equilibrium) table to find the change in concentration of each species and the equilibrium concentrations.
Using the ICE table and the equilibrium constant expression, we can solve for the equilibrium concentrations of CH4, CCl4, and CH2Cl2.
A chef is making deluxe sandwiches for a special guest. They call for 3 slices of jalapeno cheddar cheese and 5 slices of honey ham. If he has 180 slices of each in the kitchen storage, how many sandwiches can he make before one of his ingredients run out.?
Answer:
The chef can be able to make 36 sandwiches.
Explanation:
3 slices of jalapeno cheddar cheese + 5 slices of honey ham → 1 sandwich
According to information, 3 slices of jalapeno cheddar cheese will combine with 5 slices of honey ham to give 1 sandwich.
180 slices of jalapeno cheddar cheese will combine with:
[tex]\frac{5}{3}\times 180=300 [/tex] slices of honey ham
But we are having only 180 slices of honey ham. tghe number of sandwiches will depend upon number of slices of honey ham.
180 slices of honey ham will combine with:
[tex]\frac{3}{5}\times 180=108 [/tex] slices of jalapeno cheddar cheese
From 5 slices of honey ham we can make 1 sandwich,then from 180 slices of hinry ham we will be able make:
[tex]\frac{1}{5}\times 180=36 sandwiches[/tex]
The chef can be able to make 36 sandwiches.
If a mixture of total pressure 50 psia obeys Raoult's law and a species has a vapour presse of 20psia, what is the DePriester K-value for the species in the mixture?
Answer : The DePriester K-value for the species in the mixture is, 0.4
Explanation :
According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.
[tex]p_i=X_i\times p[/tex] ........(1)
According to the Raoult's law, the partial pressure exerted in gas phase by a component is equal to the product of the vapor pressure of that component and its mole fraction for an ideal liquid solution.
[tex]p_i=Y_i\times p_v[/tex] ........(2)
When the gas and the liquid are in equilibrium then these partial pressures must be the same.
[tex]X_i\times p=Y_i\times p_v[/tex]
or,
[tex]\frac{X_i}{Y_i}=\frac{p_v}{p}[/tex]
This ration is called as equilibrium ratio [tex]K_i[/tex] of the i-th component.
[tex]K_i=\frac{X_i}{Y_i}=\frac{p_v}{p}[/tex] .......(3)
As we are given that,
Total pressure = [tex]p=50psia[/tex]
The vapor pressure = [tex]p_v=20psia[/tex]
According to the relation (3), we get
[tex]K_i=\frac{X_i}{Y_i}=\frac{p_v}{p}=\frac{20}{50}=0.4[/tex]
Therefore, the DePriester K-value for the species in the mixture is, 0.4
The weak base ammonia, NH3, and the strong acid hydrochloric acid react to form the salt ammonium chloride, NH4Cl. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.289 M solution of ammonium chloride at 25∘C
Final answer:
To calculate the pH of a 0.289 M ammonium chloride solution, first determine the Ka of the ammonium ion from the Kb of ammonia and set up an ICE table to find the hydronium ion concentration. Then calculate the pH using the negative logarithm of the hydronium ion concentration.
Explanation:
The pH of a solution of ammonium chloride can be calculated by first determining the Ka (acid dissociation constant) of the ammonium ion (NH4+), which is the conjugate acid of ammonia (NH3). Given that the Kb for ammonia is 1.8×10−5, the Ka for ammonium can be calculated using the relationship Ka = Kw/Kb, where Kw is the ion-product constant for water (1.0×10−14 at 25°C). In this case, Ka = 1.0×10−14 / 1.8×10−5 = 5.6×10−10.
Knowing the Ka, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentration of hydronium ions, H3O+, produced. Since ammonium chloride is a strong electrolyte, it completely dissociates in water, thus initial [NH4+] is 0.289 M, and initial [H3O+] is 0. After the equilibrium is established, we calculate the concentration of H3O+ and subsequently find the pH of the solution. The pH is determined using the formula pH = -log[H3O+]. For a solution of ammonium chloride, this results in an acidic pH due to the formation of H3O+ ions.
The pH of a 0.289 M solution of ammonium chloride is approximately 4.89
This is calculated by determining the dissociation constant (Ka) for NH₄⁺ and using the concentration of H₃O⁺ ions to find the pH.
The Ka is found using the relation to the base dissociation constant (Kb) of ammonia.To determine the pH of a 0.289 M solution of ammonium chloride (NH₄Cl), we need to consider the dissociation of the ammonium ion (NH₄⁺) in water:
NH₄⁺ (aq) + H₂O (l) ⇔ H₃O⁺ (aq) + NH₃ (aq)The equilibrium constant for this reaction is the acid dissociation constant (Ka) of the ammonium ion.
We can calculate Ka using the relation Ka = Kw / Kb.
Given Kw = 1.0 × 10⁻¹⁴ and Kb for ammonia (NH₃) as 1.8 × 10⁻⁵, we find:
Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰Assuming the dissociation of NH₄⁺ is small, the concentration of H₃O⁺ is 'x', and using the initial concentration of NH₄⁺ (0.289 M) in the expression for Ka:
Ka = [tex]\frac{[H_{3}O^{+}][NH_{3} ] }{NH_{4} ^{+} }[/tex]5.6 × 10⁻¹⁰ = [tex]\frac{(x)(x)}{(0.289 - x) }[/tex] ≈ [tex]\frac{x^{2} }{0.289}[/tex]Simplifying for 'x', we get:
x² = 5.6 × 10⁻¹⁰ * 0.289x² = 1.62 × 10⁻¹⁰x (which is [H₃O⁺]) = [tex]\sqrt{(1.62 \times 10^{-10} )}[/tex] ≈ 1.27 × 10⁻⁵ MThe pH is then calculated as:
pH = -log[H₃O⁺] ≈ -log(1.27 × 10⁻⁵) ≈ 4.89Hence, the pH of a 0.289 M solution of ammonium chloride is approximately 4.89
Which of the following acids is the STRONGEST? The acid is followed by its Ka value. Which of the following acids is the STRONGEST? The acid is followed by its Ka value. HF, 3.5 × 10-4 HCOOH, 1.8 × 10-4 HClO2, 1.1 × 10-2 HCN, 4.9 × 10-10 HNO2, 4.6 × 10-4
Answer:
chlorous acid HClO₂
Explanation:
The Ka is the acidity constant and it tells you about the acidity strength of a compound. If the value of Ka is high the compound is a strong acid. If the value of Ka is low the compound is a weak acid.
The problem gives us the following compounds with the Ka values:
HF, 3.5 × 10⁻⁴
HCOOH, 1.8 × 10⁻⁴
HClO₂, 1.1 × 10⁻²
HCN, 4.9 × 10⁻¹⁰
HNO₂, 4.6 × 10⁻4
The chlorous acid HClO₂ have the highest Ka, 1.1 × 10⁻², so this one is the strongest acid.
The strength of an acid is determined by its Ka value. Among the given options: HF, HCOOH, HClO2, HCN, and HNO2, the strongest acid is HClO2 as it has the highest Ka value of 1.1 × 10-2.
Explanation:The acidity of a substance is determined by its Ka value, which is the acid dissociation constant. A higher Ka value indicates a stronger acid because it shows the acid dissociates more completely in solution. The substances you've listed are all acids and their respective Ka values are given. Here, the strongest acid would have the highest Ka value.
Looking at the given options: HF (Ka = 3.5 × 10-4), HCOOH (Ka = 1.8 × 10-4), HClO2 (Ka = 1.1 × 10-2), HCN (Ka = 4.9 × 10-10), and HNO2 (Ka = 4.6 × 10-4), the strongest acid based on their Ka values would be HClO2 as it has the highest Ka value of 1.1 × 10-2.
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The solubility of CO2 in water at 25°C and 1 atm is 0.034 mol/L. What is its solubility under atmospheric conditions? (The partial pressure of CO2 in air is 0.0003 atm.) Assume that CO2 obeys Henry’s law.
Hey there!:
Henry law solubility proportional to partial pressure of gas over a solvent :
for pressure of 1 atm s = 0.034
fro partial pressure of =0 .0003
Therefore :
Solubility = 0.0003 / 1 * 0.034
Solubility = 1.02 * 10⁻⁵ mol/L
Hope this helps!
Answer:
Its solubility under atmospheric conditions = [tex]1.02*10^{-5} mol/L[/tex]Explanation:
From Henry's law
c = kP
where
c = molar concentration
k = proportionality constant
P = pressure
Hence, solubility in water
[tex]k = \frac{c}{P}\\\\k = \frac{0.34}{1}\\\\k = 0.034mol/L-atm[/tex]
The solubility of [tex]CO_2[/tex] under atmospheric conditions in air is
[tex]c = kP\\\\c = 0.034 * 0.0003\\\\c = 1.02*10^{-5} mol/L[/tex]
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Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.20. If the molar absorptivity of yellow dye at 427 nm is 27400 M–1cm–1, what is the concentration of the solution?
Answer:
The concentration of the solution, [tex]C=7.2992\times 10^{-6} M[/tex]
Explanation:
The absorbance of a solution can be calculated by Beer-Lambert's law as:
[tex]A=\varepsilon Cl[/tex]
Where,
A is the absorbance of the solution
ɛ is the molar absorption coefficient ([tex]L.mol^{-1}.cm^{-1}[/tex])
C is the concentration ([tex]mol^{-1}.L^{-1}[/tex])
l is the path length of the cell in which sample is taken (cm)
Given,
A = 0.20
ɛ = 27400 [tex]M^{-1}.cm^{-1}[/tex]
l = 1 cm
Applying in the above formula for the calculation of concentration as:
[tex]A=\varepsilon Cl[/tex]
[tex]0.20= 27400\times C\times 1[/tex]
[tex]C = \frac{0.20}{27400\times 1} M[/tex]
So , concentration is:
[tex]C=7.2992\times 10^{-6} M[/tex]
The concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
BEER-LAMBERT EQUATION:
The concentration of a solution/sample measured using a spectrophotometer can be calculated using beer-lambert's equation as follows:A = εbc
Where;
ε = molar absorptivity of the yellow dye solution b = the path length of cuvettec = the concentration of the yellow dye solutionA = absorbance of the yellow dyec = A ÷ εb
According to this question, the absorbance of the yellow dye solution at 427 nm is 0.20, its molar absorptivity at 427 nm is 27400 M-¹cm-¹ and the cuvette length is 1.0cm. Hence, the concentration can be calculated as follows:c = A ÷ εb
c = 0.20 ÷ (27400 × 1)
c = 0.20 ÷ 27400
c = 7.29 × 10-⁶M
Therefore, the concentration of the yellow dye solution as obtained is 7.29 × 10-⁶M.
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Enzymes function most efficiently at the temperature of a typical cell, which is 37 degrees Celsius. What happens to enzyme function when the temperature rises? What happens to enzyme function when the temperature drops?
Answer:
At high temperature the enzyme becomes denatured.
Lower temperature the enzymes become inactive.
Explanation:
Enzymes do not work at high temperature since the temperature kills the cells. And at LOW temperature enzymes have no energy to perform their work hence becomes inactive.
A rigid tank contains 20 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psia and 90°F, respectively. Determine the amount of air added to the tank. The gas constant of air is R
Answer : The amount of air added to the tank will be, 1.2062 Kg.
Explanation :
First we have to calculate the volume of air by using ideal gas equation.
[tex]P_1V_1=\frac{m_1RT_1}{M}[/tex]
where,
[tex]P_1[/tex] = initial pressure of air = [tex]20psia=1.36atm[/tex]
conversion used : [tex]1psia=0.068046atm[/tex]
[tex]T_1[/tex] = initial temperature of air = [tex]70^oF=294.261K[/tex]
conversion used : [tex](70^oF-32)\frac{5}{9}+273.15=294.261K[/tex]
[tex]V_1[/tex] = initial volume of air = ?
[tex]m_1[/tex] = initial mass of air = [tex]20Ibm=9071.85g[/tex]
conversion used : [tex]1lbm=453.592g[/tex]
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.36atm)\times V_1=\frac{(9071.85g)\times (0.0821L.atm/mole.K)\times (294.261K)}{M}[/tex]
[tex]V_1=\frac{161150.9299}{M}L[/tex]
Now we have to calculate the final amount of air by using ideal gas equation.
[tex]P_2V_2=\frac{m_2RT_2}{M}[/tex]
where,
[tex]P_2[/tex] = final pressure of air = [tex]23.5psia=1.599atm[/tex]
[tex]T_2[/tex] = final temperature of air = [tex]90^oF=305.37K[/tex]
[tex]V_2[/tex] = final volume of air = [tex]V_1=\frac{161150.9299}{M}L[/tex]
[tex]m_2[/tex] = final mass of air = ?
R = gas constant = 0.0821 L.atm/mole.K
M = molar mass of air
Now put all the given values in the above expression, we get:
[tex](1.599atm)\times (\frac{161150.9299}{M}L)=\frac{m_2\times (0.0821L.atm/mole.K)\times (305.37K)}{M}[/tex]
[tex]m_2=10278.074g[/tex]
Now we have to calculate the amount of air added to the tank.
[tex]m_2-m_1=10278.074g-9071.85g=1206.224g=1.2062Kg[/tex]
conversion used : (1 Kg = 1000 g)
Hence, the amount of air added to the tank will be, 1.2062 Kg.
The relation between the volume, the pressure, and the temperature is PV = mRT. Then the amount of air added to the tank is 1.2062 kg.
What is thermodynamics?
It is a branch of science that deals with heat and work transfer.
A rigid tank contains 20 lbm of air at 20 psi and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psi and 90°F, respectively.
The ideal gas equation is
[tex]PV = \dfrac{mRT}{M}[/tex]
[tex]P_1 = 20 \ psi = 1.36 \ atm\\\\T_1 = 70 ^oF = 294.261\ K\\\\V_1 = \ \ ? \\\\m_1 = 20 \ lbm = 9071.85 \ kg[/tex]
R = 0.0821 L atm/mole K
M = molar mass of air
Now put all the given values in the ideal gas equation, we have
[tex]\rm V_1 = \dfrac{9071.85*0.0821*294.261}{1.36M}\\\\\\V_1 = \dfrac{161150.8299}{M}[/tex]
Now we have to calculate the final amount of air by using an ideal gas equation, we have
[tex]P_2 = 23.5 \ psi = 1.599\ atm\\\\T_2 = 90^oF = 305.37\ K\\\\V_2 = V_1\\\\m_2 = \ \ ?[/tex]
Then we have
[tex]m_2 = \dfrac{1.599M* \dfrac{161150.9299}{M}}{0.0821*305.37}\\\\m_2 = 10278.074 \ g[/tex]
Now we have to calculate the amount of air added to the tank will be
⇒ m₂ - m₁ = 10278.074 - 9071.85 = 1206.244 g = 1.12062 kg
Hence, the amount of air added to the tank is 1.2062 kg.
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The vapor pressure of water is 23.76 mm Hg at 25°C. How many grams of urea, CH4N2O, a nonvolatile, nonelectrolyte (MW = 60.10 g/mol), must be added to 238.2 grams of water to reduce the vapor pressure to 23.22 mm Hg ? water = H2O = 18.02 g/mol.
Answer:
18.700 g
Explanation:
As the urea is a nonvolatile and nonelectrolyte solute, it will reduce the vapor pressure of the solution according to:
[tex]P_{vs} =P_{w} *x_{w}[/tex]
Where [tex]P_{vs}[/tex] is the vapor pressure of the solution, [tex]P_{w}[/tex] is the vapor pressure of the pure water, and [tex]x_{w}[/tex] is the molar fraction of water. This equation applies just for that kind of solutes and at low pressures (23.76 mmHg is a low pressure).
From the equation above lets calculate the water molar fraction:
[tex]23.22mmHg=23.76mmHg*x_{w}\\ x_{w}=\frac{23.22mmHg}{23.76mmHg}=0.977[/tex]
So, the molar fraction of the urea should be: [tex]x_{urea}=1-x_{w}=0.023[/tex]
Then, calculate the average molecular weight:
[tex]M=x_{w}*MW_{w}+x_{urea}*MW_{urea}\\ M=0.977*18.02+0.023*60.10=18.989[/tex]
The molar fraction of urea is:
[tex]0.023=\frac{X urea mol}{S solution moles}=\frac{x urea grams}{238.2+x (solution grams)}*\frac{1 urea mol}{60.10 g}*\frac{18.989 solution grams}{1 solution mol}[/tex]
Solving for x,
[tex]x=18.700g[/tex]
A student performed a serial dilution on a stock solution of 1.33 M NaOH. A 2.0 mL aliquot of the stock NaOH (ms) was added to 18 mL of water to make the first dilution (m1). Next, 2.0 mL of the m1 solution was added to 18 mL of water to make the second solution (m2). The same steps were repeated for a total of 5 times. What is the final concentration of NaOH (m5)?
Answer:
[tex]\boxed{1.33 \times 10^{-5}\, \text{mol/L}}}[/tex]
Explanation:
Data:
c₀ = 1.33 mol·L⁻¹
Dilutions = 2 mL stock + 18 mL water
n = five dilutions
Calculations:
The general formula is for calculating a single dilution ratio (DR) is
[tex]DR = \dfrac{V_{i}}{V_{f}}[/tex]
For your dilutions,
[tex]V_{i} = \text{2 mL}\\V_{f} = \text{20 mL}\\DR = \dfrac{ \text{2 mL}}{\text{20 mL}} = \dfrac{1}{10}[/tex]
(Note: This is the same as a dilution factor of 10:1)
The general formula for the concentration cₙ after n identical serial dilutions is
[tex]c_{n} = c_{0}(\text{DR})^{n}[/tex]
So, after five dilutions
[tex]c_{5} =1.33 \left (\dfrac{1}{10} \right )^{5}=1.33\left ( \dfrac{1}{10^{5}} \right ) = \mathbf{1.33 \times 10^{-5}}\textbf{ mol/L}\\\\\text{The final concentration is $\boxed{\mathbf{1.33 \times 10^{-5}\, mol/L}}$}[/tex]