Answer:
3.30 × 10³ yearsExplanation:
1) Carbon-14 disintegration rate of a young (alive) tree: 0.266 per second per gram.
2) Carbon-14 disintegration rate of a wood sample prepeared from an object recovered at an archaelogical excavation (dead matter): 0.178 per second per gram.
3) Ratio of decay: A/A₀ = 0.178 / 0.266 = 0.669
4) Half-life equation: A/A₀ = (1/2)ⁿ, where n is the number of half-lives since the object died.
5) Substitute and solve for n:
0.669 = (1/2)ⁿ[tex]nlog(1/2)=log(0.669)\\ \\n=\frac{log(0.669)}{log(1/2)}\\ \\n=0.578[/tex]
That means that 0.578 half-life has elapsed since the wood with which the object was created was dead matter.
6) Convert number of half-lives to years:
0.578 half-life × 5715 years/half-life = 3,303 years = 3.30 × 10³ years.Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 25.0L tank with 4.5 mol of sulfur dioxide gas and 4.5 mol of oxygen gas at 30.°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.4 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Explanation:
Initial Concentration of sulfur dioxide = [tex][SO_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
Initial Concentration of oxygen= [tex][O_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
[tex]2SO_2+O_2\rightleftharpoons 2SO_3[/tex]
Initially (0.18 M) (0.18 M) 0
Eq'm (0.18 -2x) (0.18 -x) 2x
Equilibrium concentration of sulfur trioxide =[tex][SO_3]=2x=\frac{1.4 mol}{25 L}=0.056 M[/tex]
x = 0.028 M
Equilibrium concentration of sulfur dioxide =[tex][SO_2]'=(0.18 -2x)=0.18 - 0.056 =0.124 M[/tex]
Equilibrium concentration of oxygen=[tex][O_2]'=(0.18 -x)=0.18 - 0.028 =0.152 M[/tex]
The expression for an equilibrium constant will be :
[tex]K_c=\frac{[SO_3]^2}{[SO_2]'^2[O_2]'}[/tex]
[tex]K_c=\frac{(0.056 M)^2}{(0.124 M)^2(0.152 M)}=1.3418\approx 1.3[/tex]
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Final answer:
To calculate the concentration equilibrium constant (Kc) for the reaction of SO2 and O2 to form SO3, find the equilibrium concentrations from the initial amounts and amount at equilibrium, then apply the equilibrium expression. The resulting Kc for the reaction at the final temperature is 0.020 when rounded to two significant digits.
Explanation:
The calculation of the concentration equilibrium constant for the reaction between sulfur dioxide and oxygen to form sulfur trioxide at a certain temperature involves using the equilibrium concentrations of reactants and products. The balanced chemical equation for the reaction is:
2 SO2(g) + O2(g) = 2 SO3(g)
Given 4.5 mol of SO2 and 4.5 mol of O2 initially in a 25.0 L tank, and 1.4 mol of SO3 at equilibrium, we can calculate the change in moles during the reaction (δ) and thus the equilibrium concentrations ([SO2], [O2], and [SO3]). The concentration equilibrium constant (Kc) is then found using the expression:
Kc = ([SO3]2)/([SO2]2 × [O2])
Through stoichiometry and equilibrium concentration calculations, the concentrations are:
[SO3] = 1.4 mol / 25.0 L = 0.056 M
[SO2] = (4.5 mol - 1.4 mol) / 25.0 L = 0.124 M (since 1 mol of SO2 is consumed for every mol of SO3 produced)
[O2] = (4.5 mol - 0.7 mol) / 25.0 L = 0.152 M (since 0.5 mol of O2 is consumed for every mol of SO3 produced)
Plugging these values into the Kc expression gives:
Kc = (0.0562) / (0.1242 × 0.152) = 0.0197
The calculated Kc at the final temperature is 0.020 (rounded to two significant digits).
Calculate the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute.) Tro, Nivaldo J.. Chemistry (p. 617). Pearson Education. Kindle Edition.
Answer:
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
Explanation:
Vapor pressure of water at 25 °C ,[tex]p^o= 23.8 mmHg[/tex]
Vapor pressure of the solution = [tex]p_s[/tex]
Number moles of water in 5.50% NaCl solution.In 100 gram of solution, 94.5 g of water is present.
[tex]n_1=\frac{94.5 g}{18 g/mol}=5.25 mol[/tex]
Number moles of NaCl in 5.50% NaCl solution.5.50 g of NaCl in 100 grams of solution.
[tex]n_2=\frac{5.50 g}{58.5 g/mol}=0.09401 mol [/tex]
Mole fraction of the solute = [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.09401 mol}{0.09401 mol+5.25 mol}=0.01759[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal to the mole fraction of solute in the solution:
[tex]\frac{p^o-p_s}{p^o}=\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{23.8 mmHg - p_s}{23.8 mmHg}=0.01759[/tex]
[tex]p_s=23.38 mmHg[/tex]
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
23.38 mmHg is the vapour pressure.
The pressure enforced by the vapours when in the thermodynamical equilibrium state on the system is called the vapour pressure.
How to calculate the vapour pressure?Vapour pressure [tex](p^{\circ})[/tex] of water at [tex]25 ^{\circ} \rm C[/tex] = 23.8 mmHgVapour pressure of aqueous solution = [tex](p_{s})[/tex]Step 1: Moles of water in 5.50% NaCl solution when 100 grams of solution = 94.5 g of water
[tex]\begin{aligned}\rm n &= \dfrac{94.5 \;\rm g}{18\;\rm g/mol}\\\\\\&= 5.25\;\rm mol\end{aligned}[/tex]
Step 2: Moles of NaCl in 5.50% solution when 5.50 g NaCl in 100 grams solution then,
[tex]\begin{aligned}\rm n &= \dfrac{5.50\;\rm g}{58.5\;\rm g/mol}\\\\\\&= 0.09401\;\rm mol\end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute:
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}}\\\\\rm X_{2} &= \dfrac{0.09401\;\rm mol}{0.09401 + 5.25\;\rm mol}\\\\&= 0.0175\end{aligned}[/tex]
Step 4: Calculate the vapour pressure
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}} = \dfrac{p^{\circ}-p_{s}}{p^{\circ}}\\\\0.0175 &= \dfrac{23.38 \;\text{mmHg} -p_{s}}{23.38 \;\rm mmHg}\\\\&= 23.38\;\rm mmHg\end{aligned}[/tex]
Therefore, 23.38 mmHg is the vapour pressure.
Learn more about vapour pressure here:
https://brainly.com/question/14007496
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and heat during the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp.m 28.253 J-K1-mol1.
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
We know that,
The relation between the [tex]C_p\text{ and }C_v[/tex] for an ideal gas are :
[tex]C_p-C_v=R[/tex]
As we are given :
[tex]C_p=28.253J/K.mole[/tex]
[tex]28.253J/K.mole-C_v=8.314J/K.mole[/tex]
[tex]C_v=19.939J/K.mole[/tex]
Now we have to calculate the entropy change of the gas.
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
[tex]\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J[/tex]
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. [tex](w=-pdV)[/tex]
(C) Heat during the process will be,
[tex]q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J[/tex]
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.