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3 years ago

Question

Chemistry

2 answers:

Leya [2.2K]3 years ago

8 0

Answer:

The volume of oxygen gas at STP is 12.60 L

Explanation:

Step 1: data given

Mass of oxygen = 18.0 grams

Molar mass of oxygen = 32.0 g/mol

STp = 1atm and 273 K

Step 2: Calculate moles of oxygen

Moles O2 = mass O2 / molar mass O2

Moles O2 = 18.0 grams / 32.0 g/mol

Moles O2 = 0.5625 moles

Step 3: Calculate volume of the gas

p*V = n*R*T

⇒with p = the pressure at STP = 1.00 atm

⇒with V = the volume of the oxygen gas = TO BE DETERMINED

⇒with n = the number of moles of oxygen gas = 0.5625 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature at STP = 273 K

V = (n*R*T)/p

V = (0.5625*0.08206 * 273) / 1 atm

V = 12.60 L

The volume of oxygen gas at STP is 12.60 L

VMariaS [17]3 years ago

7 0

Answer:

12.62 L

Explanation:

First, we have to calculate the moles corresponding to 18.0 g of oxygen gas (MW 32.0).

18.0 g × (1 mol/32.0 g) = 0.563 mol

Then, we can find the volume occupied by 0.563 moles of oxygen at STP (273,15 K, 1.00 atm) using the ideal gas law.

P × V = n × R × T

V = n × R × T / P

V = 0.563 mol × 0.0821 atm.L/mol.K × 273.15 K / 1.00 atm

V = 12.62 L

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let the rate of diffusion of H2(r1) be x and⁷ the rate of diffusion of gas(r2) be x+x/6=7x/6.

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Now, from the Ghram's law of diffusion of gas,

$\frac{r1}{r2} = \sqrt{ \frac{m2}{m1} } \\ or \: \frac{x}{ \frac{7x}{6} } = \sqrt{ \frac{m2}{2} } \\ or \: \frac{6}{7} = \sqrt{ \frac{m2}{2} } \\ squaring \: in \: both \: sides \\ \frac{36}{49} = \frac{m2}{2} \\ or m2 = 1 .469$

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8 0

4 years ago

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olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

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