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3 years ago

What is the density of CO2 at a pressure of 0.0079 atm and 227 K? (These are the approximate atmospheric in Mars)

Chemistry

1 answer:

Strike441 [17]3 years ago

6 0

Answer:

The density is 0.0187 g/L

Explanation:

First thing to do here is to calculate the Volume of 1 mole of CO2 using the ideal gas equation

Mathematically;

PV = nRT

thus V = nRT/P

what we have are;

n = 1 mole

R is the molar has constant = 0.082 L•atm•mol^-1•K^-1

P is the pressure = 0.0079 atm

T is temperature = 227 K

Substituting these values, we have;

V = nRT/P = (1 * 0.082 * 227)/0.0079

V = 2,356.20 dm^3

This means according to the parameters given in the question, the volume of 1 mole of carbon iv oxide is 2,356.20 dm^3

But this is not what we want to calculate

What we want to calculate is the density

Mathematically, we can calculate the density using the formula below;

density = molar mass/molar volume

Kindly recall that the molar mass of carbon iv oxide is 44 g/mol

Thus the density = 44/2356.20 = 0.018674136321195 which is approximately 0.0187 g/L

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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

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