Final answer:
The mass percent of ethanol in wine is approximately 10.38%, and the molality of ethanol in wine is approximately 2.51 mol/kg. These values are calculated from the given volume percent and the density of ethanol by first determining the mass of ethanol and then the number of moles of ethanol.
Explanation:
Calculating Mass Percent and Molality of Ethanol in Wine
We are given that wine contains 12.8% ethanol by volume, and the density of ethanol is 0.789 g/cm3. To find the mass percent and molality of ethanol in wine, we proceed with the following steps:
Calculate the volume of ethanol in a 100 mL sample of wine. Since the wine is 12.8% ethanol by volume, we have 12.8 mL of ethanol.Using the density of ethanol, compute the mass of ethanol. Mass = Volume x Density, which gives us 12.8 mL x 0.789 g/mL = 10.0992 g of ethanol.The mass percent of ethanol is then calculated by comparing this mass to the total mass of the wine, assuming the rest is water (density 1.00 g/mL). If we have 87.2 mL of water, the mass of water is 87.2 g, and the total mass of the solution is 10.0992 g (ethanol) + 87.2 g (water) = 97.2992 g. Thus, the mass percent of ethanol is (10.0992 g / 97.2992 g) x 100% = approximately 10.38%.To calculate molality, we first find the moles of ethanol using its molar mass (46.07 g/mol): Moles of ethanol = 10.0992 g / 46.07 g/mol = approximately 0.219 moles of ethanol.Molality is moles of solute (ethanol) per kilogram of solvent (water). With 0.0872 kg of water, the molality is 0.219 moles / 0.0872 kg = approximately 2.51 mol/kg.Through these calculations, we have determined both the mass percent and molality of ethanol in the wine sample.
Calculate the linear momentum of photons of wavelength
(a) 600 nm,
(b) 70 pm,
(c) 200 m.
Answer:
a) p = 1.10 * 10⁻²⁷ kg·m/s
b) p = 9.46 * 10⁻²⁴ kg·m/s
c) p = 3.31 * 10⁻³⁶ kg·m/s
Explanation:
To solve this problem we use the de Broglie's equation, which describes the wavelenght of a photon with its momentum:
λ=h/p
Where λ is the wavelength, h is Planck's constant (6.626 * 10⁻³⁴ J·s), and p is the linear momentum of the photon.
Rearrange the equation in order to solve for p:
p=h/λ
And now we proceed to calculate, keeping in mind the SI units:
a) 600 nm= 600 * 10⁻⁹ m
p=(6.626 * 10⁻³⁴ J·s) / (600*10⁻⁹m) = 1.10 * 10⁻²⁷ kg·m/s
b) 70 pm= 70 * 10⁻¹² m
p=(6.626 * 10⁻³⁴ J·s) / (70*10⁻¹²m) = 9.46 * 10⁻²⁴ kg·m/s
c) 200 m
p=(6.626 * 10⁻³⁴ J·s) / (200m) = 3.31 * 10⁻³⁶ kg·m/s
Suppose that NaCl is added to hexane (C6H14) instead of water. Which of the following intermolecular forces will exist in the system?
Check all that apply.
Ion-dipole force between Na+ ions and a hexane molecule
Ion-ion force between Na+ and Cl− ions
Dipole-dipole force between two hexane molecules
Hydrogen bonding between Na+ ions and a hexane molecule
London dispersion force between two hexane molecules
Answer:
Ion-ion force between Na+ and Cl− ions
London dispersion force between two hexane molecules
Explanation:
"Ion-dipole force between Na+ ions and a hexane molecule " does not exist since hexane has only non-polar bonds and therefore no dipole.
"Ion-ion force between Na+ and Cl− ions " exists since both are ions.
"Dipole-dipole force between two hexane molecules " does not exist since hexane molecules do not have a dipole.
"Hydrogen bonding between Na+ ions and a hexane molecule " does not exist since the hydrogen in the hydrogen bond must be bonded directly to an electronegative atom, which hexane does not have since it is a hydrocarbon.
"London dispersion force between two hexane molecules" exist since hexane is a molecular compound.
Answer:
Ion-ion force between Na⁺ and Cl⁻ ions.
London dispersion force between two hexane molecules.
Explanation:
Suppose that NaCl is added to hexane (C₆H₁₄) instead of water. Which of the following intermolecular forces will exist in the system?
Check all that apply.
Ion-dipole force between Na⁺ ions and a hexane molecule. NO. Hexane is nonpolar so it can interact through ion-dipole forces.Ion-ion force between Na⁺ and Cl⁻ ions. YES. Na⁺ is a cation (positive ion) and interacts with the anion (negative ion) Cl⁻ through ion-ion forces.Dipole-dipole force between two hexane molecules. NO. Since hexane molecules are not dipoles, they do not show dipole-dipole interaction.Hydrogen bonding between Na⁺ ions and a hexane molecule. NO. Hydrogen bonding requires H bonded to a very electronegative atom, such as N, O or F, which does not happen in hexane.London dispersion force between two hexane molecules. YES. The kind of interaction between nonpolar molecules is London dispersion force.Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O 16 Δ[H2O]Δt 14 Δ[NO2]Δt - 14 Δ[NH3]Δt - 17 Δ[O2]Δt All of the above are valid expressions of the reaction rate.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
[tex]4NH_3+7O_2\rightarrow 4NO_2+6H_2O[/tex]
The expression for rate of reaction for the reactant :
[tex]\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}[/tex]
[tex]\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}[/tex]
The expression for rate of reaction for the product :
[tex]\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}[/tex]
[tex]\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}[/tex]
From this we conclude that, all the options are correct.
Based on the equation of the reaction, all of the given expressions are valid expressions of the reaction rate.
What is the rate of a chemical reaction?The rate of a chemical reaction is the rate at which reactants are used up or the rate at which products are formed.
The rate of a reaction is obtained from the balanced equation of the reaction.
The equation of the reaction is given below:
4NH3 + 7O2 → 4NO2 + 6H2ORate of disappearance of NH3 = - 1/4 × Δ[NH3]/Δt
Rate of disappearance of O2 = - 1/7 × Δ[O2]/Δt
Rate of formation of NO2 = 1/4 × Δ[NO2]/Δt
Rate of formation of H2O = 1/6 × Δ[H2O]/Δt
Therefore, all of the above are valid expressions of the reaction rate.
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A solution is prepared by mixing 30 ml of a 0.2 M hydrochloric acid (HCl) with 20 ml of a 0.3 M potassium hydroxide solution (KOH), what will be the final pH? (Show your calculation to get full credit.) Hint: Determine whether hydrochloric acid and potassium hydroxide are strong or weak acids or bases.
Answer:
The final pH is 7
Explanation:
KOH + HCl ------> KCl + H2O
KOH ----> K+ + OH-
0,3 M 0,3 M + 0,3 M
HCl -----> H+ + Cl-
0,2 M 0,2M + 0,2M
In both cases they are strong acid and base because they dissociate completely; The initial molarity is the same, at the end of the dissociation.
As we have an acid next to a strong base, the final solution will have a neutralization reaction where the OH- of the base react with the H + of the acid to give water. The excess of OH- or H + defines whether the solution will have an acid or alkaline pH.
Molarity = moles / volume (L)
Moles of OH- = 0,3 M x 0,020 L = 0,006 moles
(remember volume in L, so 20mL = 0,020 L)
Moles of H+ = 0,2 M x 0,030 L = 0,006 moles
As we have the same quantity for OH- and H+, the solution will have neutral pH.
What is the wavelength of radiation that has a frequency
of6.912 x 10-14 s-1?
Answer: The wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]
Explanation:
The relationship between wavelength of light and frequency of light is given by the equation:
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of radiation = [tex]6.912\times 10^{-14}s^{-1}[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength of radiation = ?
Putting values in above equation, we get:
[tex]6.912\times 10^{-14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=4.34\times 10^{21}m[/tex]
Hence, the wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]
You opened up a new 500. mL bottle of RobitussinTM. The dosage is 1.1 tablespoon (TBS).
How many doses are in the bottle?
1 tsp. = 4.93 mL
3 tsp. = 1 TBS.
b)What is the mass of 35.2 mL of a saline (salt water) solution that has a density of 1.12 g cm−3?
Answer:
30.73 Doses and 39.42 g
Explanation:
It is necessary to know the conversion factors in that manner change the initial 500 mL to the number of doses as follow:
[tex]500 ml * \frac{1tsp}{4.93mL} * \frac{1TBS}{3tsp} * \frac{1 Dose }{1.1 TBS} = 30.7 doses\\[/tex]
And the for b. it is such necessary to take in account that 1 mL = 1 [tex]cm^{3}[/tex]
35.2 mL * 1.12 [tex]\frac{g}{cm^{3} }[/tex] = 39.42 g
How many grams of 48.0 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.11 M NaOH? (Enter your answer using two digits.)
Answer:
The answer is 916.67 g
Explanation:
48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:
[tex]\frac{48 g NaOH}{100 g solution}[/tex] x [tex]\frac{1 mol NaOH}{40 g}[/tex] = 0.012 mol NaOH/100 g solution
Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.
0.012 mol NaOH ------------ 100 g solution
0.11 mol NaOH------------------------- X
X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g
We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.
One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer in L using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
28.52 L
Explanation:
First, let's calculate the density of the ocean, which is the mass divided by the volume:
d = m/V
d = 35.06/1
d = 35.06 g/L
So, for a mass of 1.00 kg = 1000.00 g
d = m/V
35.06 = 1000.00/V
V = 1000.00/35.06
V = 28.52 L
How all the data are expressed with two significant figures, the volume must also be expressed with two.
Hydrocyanic acid has a Ka of 4.0
x10-10. What is the percent ofionization of
a 1.0 molar solution?
Answer:
0.002 %
Explanation:
Given that:
[tex]K_{a}=4.0\times 10^{-10}[/tex]
Concentration = 1.0 M
Consider the ICE take for the dissociation of Hydrocyanic acid as:
HCN ⇄ H⁺ + CN⁻
At t=0 1.0 - -
At t =equilibrium (1.0-x) x x
The expression for dissociation constant of Hydrocyanic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]
[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]
x is very small, so (1.0 - x) ≅ 1.0
Solving for x, we get:
x = 2×10⁻⁵ M
Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]
Recognize Physical and Chemical Properties of Alcohols Question Which of the following statements about the -OH group of alcohols is NOT true? Select the correct answer below: O The hydrogen has a slight positive charge. O The hydrogen is slightly acidic. O The hydrogen cannot participate in hydrogen bonding with other molecules. O none of the above FEEDBACK MORE INSTRUCTION SUBMIT Content attribution
Answer:
The hydrogen cannot participate in hydrogen bonding with other molecules.
Explanation:
Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.
Thus, hydrogen bonding is present in alcohols as the oxygen atom is linked to hydrogen
Partially positive end of the hydrogen atom is attracted to partially negative end of the oxygen atom. It is strong force of attraction between the molecules. Thus, hydrogen acquires slight positive charge and become electron deficient or acidic.
Hence, option C is incorrect.
In a particular experiment at 300 ∘C, [NO2] drops from 0.0100 to 0.00800 M in 100 s. The rate of appearance of O2 for this period is ________ M/s. In a particular experiment at 300 , drops from 0.0100 to 0.00800 in 100 . The rate of appearance of for this period is ________ . 4.0×10−3 2.0×10−3 2.0×10−5 4.0×10−5 1.0×10−5
Answer:
[tex]-r_{O_2}=1\times{10}^{-5}\frac{M}{s}[/tex]
Explanation:
The decomposition of [tex]NO_2[/tex] follows the equation
[tex]2NO_2\rightarrow2NO+O_2[/tex]
By definition, the rate of a chemical reaction can be expressed by
[tex]-r_{NO_2}=\frac{d\left[NO_2\right]}{dt}=\frac{0.010-0.008\ M}{100\ s}=2\times{10}^{-5}\frac{M}{s}[/tex]
The rate of appearance of [tex]O_2[/tex] is related to the rate of disappearance of [tex]NO_2[/tex] by the stoichiometry. This means that, for each mole of [tex]O_2[/tex] that appears 2 moles of [tex]NO_2[/tex] are consumed. So
[tex]-r_{O_2}=-r_{NO_2}\times\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1\times{10}^{-5}\frac{M}{s}[/tex]
The rate of appearance of O2 for this period is -1.0x10^-5 M/s.
Explanation:The rate of appearance of O2 for a given period can be determined using the change in concentration of NO2 over that period. In this case, the concentration of NO2 drops from 0.0100 to 0.00800 M in 100 s. First, calculate the change in concentration of NO2:
Δ[NO2] = [NO2]final - [NO2]initial = 0.00800 M - 0.0100 M = -0.00200 M
Since the reaction is 2NO2 → 2NO + O2, the stoichiometric ratio between NO2 and O2 is 2:1. Therefore, the change in concentration of O2 is half of the change in concentration of NO2:
Δ[O2] = -0.00200 M ÷ 2 = -0.00100 M
Finally, divide the change in concentration of O2 by the time period to find the rate of appearance of O2:
Rate = Δ[O2] ÷ time = -0.00100 M ÷ 100 s = -1.0x10-5 M/s
A sample of the compound MSO4 weighing 0.1131 g reacts with BaCl2 and yields 0.2193 g BaSO4. What is the elemental mass of M and its identity? Hint: All the SO42- from the MSO4 appears in the BaSO4
Answer:
The atomic mass of the given metal M is 24.3 g/mol.
Therefore, the given metal M is magnesium (Mg)
Explanation:
Reaction involved: MSO₄ + BaCl₂ → BaSO₄ + MCl₂
Atomic mass (g/mol): oxygen (O)=16, sulphur (S)=32,
Molar mass of SO₄²⁻ = 96 g/mol and molar mass of BaSO₄ = 233.38 g/mol
Let the atomic mass of M be m g/mol.
Therefore, molar mass of MSO₄= (m + 96) g/mol
If 0.1131 g of MSO₄ gives 0.2193 g of BaSO₄ on reaction
Then, (m + 96) g/mol of MSO₄ gives 233.38 g/mol of BaSO₄
Therefore, (m + 96) g/mol of MSO₄ = [(233.38 g/mol) × (0.1131 g)] ÷ (0.2193 g)
⇒(m + 96) g/mol = [26.395] ÷ (0.2193)
⇒(m + 96) g/mol = 120.36 g/mol
⇒m = 120.36 g/mol - 96 g/mol = 24.36 g/mol ≈ 24.3 g/mol
Since, the atomic mass of the given metal M is 24.3 g/mol.
Therefore, the given metal M is magnesium (Mg)
A patient is undergoing a blood test to determine her cholesterol levels. The sample tube holds 3.0 cc (cubic centimeters). What is this volume in liters?
Answer:
The sample tube holds 3.0 cc which is equal to 0.0030 Liters.
Explanation:
Volume of the blood sample in the sample tube = 3.0 cc
cc = cubic centimeter ; A unit to measure a volume
1 cubic centimeter is equal to 0.001 Liters of volume.
[tex]1 cm^3=0.001 L[/tex]
So in 3.0 cubic centimeter there will be:
[tex]3.0 cm^3=3.0\times 0.001 L=0.0030 L[/tex]
The sample tube holds 3.0 cc which is equal to 0.0030 Liters.
in carbon tetrachloride proceeds as follows: 2N2O5→4NO2+O2. The rate law is first order in N2O5. At 64 ∘C the rate constant is 4.82 ×10−3s−1. Part A Part complete Write the rate law for the reaction. Write the rate law for the reaction. rate=2.41×10−3s−1[N2O5] rate=4.82×10−3s−1[N2O5] rate=9.64×10−3s−1[N2O5] rate=4.82×10−3s−1[N2O5]2 Previous Answers Correct Part B What is the rate of reaction when [N2O5]= 2.80×10−2 M ?
Answer:
Part A
[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]
Part B
[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]
Explanation:
Part A
The rate law is the equation that relates the rate of the reaction, the kinetic constant and the concentration of the reactant or reactants.
For the given chemical reaction we can write a general expression for the rate law as follows:
[tex]rate= k * [N2O5]^{x}[/tex]
where k is the rate constant and x is the order of the reaction with respect of N2O5 concentration. Particularly, a first order reaction kinetics indicate that the rate of the reaction is directly proportional to the concentration of only one reactant. Then x must be 1.
Replacing the value of the rate constant given in the text we can arrive to the following expression for the rate law:
[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]
Part B
Replacing the value of the concentration of N2O5 given, we can get the rate of reaction:
[tex]rate= 4.82*10^{-3}s^{-1} *2.80*10^{-2}M[/tex]
[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]
QUESTION 1 Choose the statement that best reflects the three basic steps of any recrystallization Dissolve the impure solid in a minimal amount of boiling solvent: cool the solution to form crystals, vacuum filter the solution to collect the pure crystals. Dissolve the impure solid in an excess of solvent at room temperature; cool the solution to form crystals: vacuum filter to collect the pure crystals O Dissolve the impure solid in an excess of boiling solvent: cool the solution to form crystals vacuum filter to collect the pure crystals. Dissolve the impure solid in a minimal amount of solvent at room temperature; cool the solution to form crystals, vacuum filter to collect the pure crystals.
Answer:
Dissolve the impure solid in a minimal amount of boiling solvent, cool the solution to form crystals, vacuum filter the solution to collect the pure crystals.
Explanation:
Recrystallization is a process when a solid with impurities is purified. To do this a solvent of the compound we want must be used. We need to use only the quantity necessary to dissolve the compound of interest, otherwise, the solvent will dissolve the impurities or it will interfere in the crystallization.
For most of the solids, the solubility increases with the increase of the temperature, so to speed up the process, heat must be added at the system, or the solvent must be boiling. Then, the solution will be cooled to form the crystals of the compound purified, and then it must be filtered in a vacuum because the crystals can slow down the filtration.
Calculate the molarity of a solution made by
dissolving0.145mol Na2SO4 in enough water to form exactly 750mL
ofsolution.
Answer:
0.1933 M
Explanation:
Data provided in the question:
Number of Moles of Na₂SO₄ initially taken = 0.145 mol
Volume of solution, V = 750 mL = 0.750 L
Now, the molarity is given as:
Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]
or
Molarity = [tex]\frac{0.145}{0.750}[/tex]
or
Molarity = 0.1933 M
The molarity of the solution is 0.1933 M.
Explanation:To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.
So, Molarity = 0.145 mol / 0.75 L = 0.1933 M
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prob.: Consider the combustion of butane (C4H10):
2C4H10(g) + 13O2(g) ==> 8CO2(g) + 10 H2O(L)
IN A PARTICULAR REACTION, 5.0 MOLES OF C4H10 ARE REACTED
WITHAN EXCESS OF O2. CALCULATE THE NUMBER OF MOLES OF CO2
FORMED.
When 5.0 moles of C₄H₁₀ react with an excess of O2, 20.0 moles of CO₂ are formed.
According to the balanced chemical equation:
2C₄H₁₀(g) + 13O₂(g) ==> 8CO₂(g) + 10H₂O(L)
The stoichiometric coefficient of CO₂ is 8. This means that for every 2 moles of C₄H₁₀ reacted, 8 moles of CO₂ are formed. Therefore, when 5.0 moles of C₄H₁₀ react, the number of moles of CO₂ formed can be calculated using the stoichiometric ratio:
Number of moles of CO₂ = (5.0 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀ )
= 20.0 moles CO₂
So, when 5.0 moles of C₄H₁₀ react with an excess of O₂, 20.0 moles of CO₂ are formed.
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In the combustion of butane, 2 moles of C4H10 react to produce 8 moles of CO2, making a 1:4 ratio. Therefore, when 5.0 moles of butane reacts it produces 20 moles of CO2 due to this stoichiometric ratio.
Explanation:In the balanced chemical equation for the combustion of butane, we see that '2 moles of C4H10' react to produce '8 moles of CO2'. This is a ratio of 1:4. So, for each mole of butane that reacts, 4 moles of carbon dioxide are produced.
If in a particular reaction, we have '5.0 moles of C4H10', then according to the stoichiometric ratio (1:4) we can calculate the number of moles of CO2 formed. By multiplying the moles of C4H10 by 4, i.e., 5.0 x 4, we get '20 moles of CO2'. Therefore, when 5.0 moles of butane reacts with excess oxygen, 20 moles of CO2 are produced.
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A culture of E.coli bacteria doubles every 40 minutes. If there are 50 bacteria initially, how many will there be after 12 hours? Round your answer to the nearest whole number
Answer:
There will be 13107200 number of bacteria after 12 hours
Explanation:
1 hour = 60 minutes
So, 12 hours = ([tex]12\times 60[/tex]) minutes = 720 minutes
Initially there are 50 bacteria.
As number of bacteria doubles in every 40 minutes therefore rate of increase in number of bacteria will be similar to a first order reaction.
Hence, number of bacteria after 720 minutes = [tex]50\times (2)^{\frac{720}{40}}[/tex] = 13107200
it is not from thermodynamic energy?
a) utilization
b) kinetic
c) potential
d) stored energy
Answer:
(a) Utilization
Explanation:
Thermodynamic energy are:
(1) Kinetic energy:Kinetic energy is that type of energy of a body which occurs due to the motion of body. Kinetic energy is always positive. In a bound system, the system remains bound as long as the kinetic energy is less than the potential energy due to the interaction of the body.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
(2) Potential energy:Potential energy is defined as the energy in which the energy is possessed by a body or a system for doing some work, by virtue of its position above the ground level.
Therefore,
Potential energy = PE = mgh
This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 m s⁻².
(3) Stored energy:Stored energy is a Potential energy.
Hence, the only energy which is not a thermodynamic energy is option (a) Utilization energy.
Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.
Answer: Option (c) is the correct answer.
Explanation:
Entropy is defined as the degree of randomness that is present within the particles of a substance.
As [tex]NH_{4}NO_{3}[/tex] is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ([tex]NH^}{+}_{4}[/tex]) and nitrate ions ([tex]NO^{-}_{3}[/tex]).
Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.
Thus, we can conclude that ammonium nitrate ([tex]NH_{4}NO_{3}[/tex]) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.
A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 98.4 , 104.3 , 97.4 , 106.7 , and 93.0 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor.
Explanation:
Actual value = 102.0 mg/dL
Calculation of absolute error and relative error for the measured value, 98.4 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 98.4 | = 3.6
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{3.6}{102.0} = 0.03529[/tex]
Calculation of absolute error and relative error for the measured value, 104.3 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 104.3 | = 2.3
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{2.3}{102.0} = 0.02255[/tex]
Calculation of absolute error and relative error for the measured value, 97.4 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 97.4 | = 4.6
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{4.6}{102.0} = 0.04509[/tex]
Calculation of absolute error and relative error for the measured value, 106.7 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 106.7 | = 4.7
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{4.7}{102.0} = 0.04607[/tex]
Calculation of absolute error and relative error for the measured value, 93.0 mg/dL
Absolute error = |Actual value - measured value|
= | 102.0 - 93 | = 9
Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]
= [tex]\frac{9}{102.0} = 0.08823[/tex]
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.379 L flask at 1191 K. At equilibrium, the flask contains 0.145 mol of CO gas, 0.218 mol of H2 gas, and 0.25 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1191 K)?
Enter to 4 decimal places.
HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.
Answer: The concentration of water at equilibrium is 0.3677 mol/L
Explanation:
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by [tex]K_{c}[/tex]
For a general chemical reaction:
[tex]aA+bB\rightarrow cC+dD[/tex]
The [tex]K_{c}[/tex] is written as:
[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]
The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:
[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]
The [tex]K_{c}[/tex] is represented as:
[tex]K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex] ....(1)
To calculate the concentration, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
[tex][CO]=\frac{0.145mol}{0.379L}=0.383mol/L[/tex]
[tex][H_2]=\frac{0.218mol}{0.379L}=0.575mol/L[/tex]
[tex][CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L[/tex]
[tex]K_c=0.30[/tex]
Putting values in equation 1, we get:
[tex]0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}[/tex]
[tex][H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677[/tex]
Hence, the concentration of water at equilibrium is 0.3677 mol/L
Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling point of the solvent
Pure dissolved in 0.284 ° C. Determine the Kb value of solvent B. Set the molecular weight of A = 106.
Answer:
2.1 °C/m
Explanation:
Hello, for this exercise, consider the formula:
[tex]T_{solution}-T{solvent}=K_bm_solute[/tex]
Considering that the difference in the temperature is 0.284°C, and the given molality by:
[tex]m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m[/tex]
Now, solving for [tex]K_b[/tex], we get:
[tex]K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m[/tex]
Best regards.
Which of the following reactions is a double displacement reaction? A) HCI (9) - Hz (g) + Cl2 (9) B) HCI (aq) + NaOH (aq) - H2O (1) + Naci (aq) C) Mg () + 2HCl (aq) - MgCl2 (aq) + H2(g) D) HCI (9) + C-H10 (9) -CsH..CI (9)
Answer:
HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)
Explanation:
A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.
Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume
ik its not B
Answer:
option D= temperature and volume
Explanation:
Definition:
Charle's law stated that ,
" At constant pressure, the volume of given amount of gas is directly proportional to its absolute temperature"
V∝ T
V= kT
OR
V/T = k (k is proportionality constant)
if the temperature is changed from T1 to T2 then the volume of gas is also changed from V1 to V2. Then expression will be:
V1/T1= k and V2/T2=k
V1/T1=V2/T2
suppose a cylinder is filled with a gas having volume V1 at temperature T1. When the gas is heated its temperature raises from T1 to T2 and its volume also increased with increase of temperature from V1 TO V2.
In Charles' law, the two changing properties are temperature and volume, with pressure maintained constant. The law indicates a direct proportionality between temperature and volume for a given amount of gas.
Explanation:The two changing properties in Charles' law are temperature and volume. Charles' law states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. This means if the temperature of the gas increases, its volume increases as well, provided the pressure and the amount of gas remain constant. In other words, doubling the absolute temperature at constant pressure will double the volume. It is important to note that in gas laws, temperatures must always be expressed in kelvins.
What is the chemistry principle for silly putty and candle
waxas it pertains to hydrophilic/hydrophobic, polar/nonpolar
andsoluble/non-soluble and binding/non-binding?
Explanation:
Silly Putty -
It is an inorganic polymer , in which the molecules are linked via covalent bonding . And the bonds can easily be broken , by the application of even a small amount of pressure .
The composition of silly putty is , the Elmer's glue solution and the sodium borate in the ration , 4 : 1 .
silly putty is polar , hydrophilic and binding in nature .
Candle Wax -
The Wax of the candle is an ester of the ethylene glycol and fatty acids , and the chemical properties of was is , it is hydrophobic in nature and is insoluble in water as well .
The principles affecting the properties of silly putty and candle wax involve their hydrophobic, nonpolar nature, making them insoluble in water and more soluble in nonpolar solvents like heptane. Solubility depends on the balance of intermolecular forces and molecule polarity, with polar and ionic substances being hydrophilic. Heat transfer during solvation is dictated by the relative strength of solute-solvent interactions compared to original solute-solute and solvent-solvent interactions.
The chemistry principle for substances such as silly putty and candle wax as it relates to hydrophilic/hydrophobic, polar/nonpolar, and soluble/non-soluble properties is rooted in their intermolecular forces and molecular structure. Silly putty, which is a silicone polymer, is primarily hydrophobic and non-polar, making it insoluble in water and more soluble in nonpolar solvents. On the other hand, candle wax, typically composed of long-chain hydrocarbons, is also nonpolar and hydrophobic, displaying similar solubility characteristics. Things that are polar and can dissolve in water are referred to as being hydrophilic, which allows them to form hydrogen bonds with water. This property is not present in nonpolar substances like silly putty and candle wax, which resist mixing with water.
In terms of solubility predictions for various substances:
Vegetable oil is nonpolar and more soluble in a nonpolar solvent like heptane.
Isopropyl alcohol is polar due to the presence of an -OH group and more soluble in water.
Potassium bromide is ionic and highly soluble in water which is a polar solvent.
When solutions form, heat may be released or absorbed depending on the intermolecular forces involved. In solvation where the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions, heat is released (exothermic). Conversely, if breaking the original interactions requires more energy than is released during new interaction formation, the process is endothermic and absorbs heat.
1 kg of Oxygen gas is present in a 500 L tank. Find the specific and molar volumes of the gas.
Answer:
specific volume O2 = 0.5 Kg/m³
molar volume O2 = 8 E-3 m³/mol
Explanation:
specific volume (Sv):
∴ Sv = 1 / ρ
∴ ρ = mass / volume
∴ V = 500 L * ( m³/1000 L) = 0.5 m³
∴ ρ = 1 Kg / 0.5m³ = 2 Kg/m³
⇒ Sv = 1 / 2 Kg/m³ = 0.5 m³/Kg
molar volume ( Vm ):
∴ Vm = volume/mol
∴ Mw O2 = 1000g O2 * ( mol/16g O2) = 62.5 mol O2
⇒ Vm = 0.5 m³ / 62.5 mol
⇒ Vm = 8 E-3 m³/mol
What is the unit of catalytic reaction rate?
a) Mole/time
b) Mole * kg * time
c) Mole * kg / time
d) other
Answer:
The answer is the option "A"
Explanation:
The reaction rate is the amount of substance formed or transformed per unit of time. The catalytic action allows to increase the reaction rate or the selectivity. The units are moles / time (mol / s)
mol = moles
s = seconds
So the correct answer is "A"
Calculate the cell potential for a cell operating with the following reaction at 25 degrees Celsius, in which [MnO4^1-] = .01M, [Br^1-] = .01M, [Mn^2+] = .15M, and [H^1+] = 1M. The reaction is 2 MnO4^1-(aq) + 10 Br^1-(aq) + 16 H^1+(aq) --> 2 Mn^2+(aq) + 5 Br2(l) + 8 H2O(l)
Answer: The cell potential of the cell is 0.31 V
Explanation:
We know that:
[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.
Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex] ( × 5)
Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex] ( × 2)
The net reaction follows:
[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=1.51-(1.07)=0.44V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.44 V
n = number of electrons exchanged = 10
[tex][H^{+}]=1M[/tex]
[tex][Mn^{2+}]=0.15M[/tex]
[tex][MnO_4^{-}]=0.01M[/tex]
[tex][Br^{-}]=0.01M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]
Hence, the cell potential of the cell is 0.31 V
For the reaction A+B+C=>D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:
Trial A(M) --------B(M)------ C(M) ------ Initial Rate(M/s)
1 ------- 0.40 -------- 0.40 ------ 0.40 --------- 1.2 x 10^-4
2 --------0.40 --------- 0.40 -------1.20 -----------3.6 x 10^-4
3------- 0.80 ----------0.40------- 0.40 ---------- 4.8 x 10^-4
4 ------- 0.80 ----------0.80 -------0.40 ----------4.8 x 10^-4
What is the value of the rate constant k for this reaction?
Answer : The value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]A+B+C\rightarrow D+E[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]
where,
a = order with respect to A
b = order with respect to B
c = order with respect to C
Expression for rate law for first observation:
[tex]1.24\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c[/tex] ....(1)
Expression for rate law for second observation:
[tex]3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c[/tex] ....(2)
Expression for rate law for third observation:
[tex]4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c[/tex] ....(3)
Expression for rate law for fourth observation:
[tex]4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c[/tex] ....(4)
Dividing 1 from 2, we get:
[tex]\frac{3.6\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1[/tex]
Dividing 1 from 3, we get:
[tex]\frac{4.8\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2[/tex]
Dividing 3 from 4, we get:
[tex]\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]1.24\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1[/tex]
[tex]k=1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]
Hence, the value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]