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tino4ka555 [31]
3 years ago
7

A sample of 88.0 g CO2 is held at 291 K in a 40.0 L container. What is the pressure this gas exerts on the container. Express yo

ur answer in kPa.
Chemistry
1 answer:
SashulF [63]3 years ago
7 0

Answer:- 121 kPa.

Solution:- mass, temperature and volume are given for carbon dioxide gas asked to calculate the pressure this gas exerts on the container.

It is based on ideal gas law equation, PV = nRT .

Where, P is the pressure in atm, V is the volume in Liters, n is the number of moles of the gas, R is the universal gas constant and it's value is \frac{0.0821atm.L}{mol.K} and T is the kelvin temperature.

Given data:- mass = 88.0 g

T = 291 K

V = 40.0 L

P = ?

We need to convert the mass to moles and for this we divide the mass by molar mass. Molar mass of carbon dioxide gas is 44.01 gram per mol.

n=88.0g(\frac{1mol}{44.01g})

n = 2.00 mol

The equation could be rearranged for pressure as:

P=\frac{nRT}{V}

Let's plug in the values and solve this for P.

P=\frac{2.00mol*(\frac{0.0821atm.L}{mol.K})*291K}{40.0L}

P = 1.1946 atm

Since the answer is asked to report in kPa, let's convert atm to kPa.

1 atm = 101.325 kPa

So, 1.1946atm(\frac{101.325kPa}{1atm})

= 121 kPa

So, the pressure this gas exerts on the container is 121 kPa.

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<u>Answer:</u> The percentage abundance of _{29}^{63}\textrm{Cu} and _{29}^{65}\textrm{Cu} isotopes are 75.77% and 24.23% respectively

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Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{29}^{63}\textrm{Cu} isotope be 'x'. So, fractional abundance of _{29}^{63}\textrm{Cu} isotope will be '1 - x'

  • <u>For _{29}^{63}\textrm{Cu} isotope:</u>

Mass of _{29}^{63}\textrm{Cu} isotope = 62.9396 amu

Fractional abundance of _{29}^{63}\textrm{Cu} isotope = x

  • <u>For _{29}^{65}\textrm{Cu} isotope:</u>

Mass of _{29}^{65}\textrm{Cu} isotope = 64.9278 amu

Fractional abundance of _{29}^{65}\textrm{Cu} isotope = 1 - x

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Putting values in equation 1, we get:

63.546=[(62.9396\times x)+(64.9278\times (1-x))]\\\\x=0.6950

Percentage abundance of _{29}^{63}\textrm{Cu} isotope = 0.6950\times 100=69.50\%

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Hence, the percentage abundance of _{29}^{63}\textrm{Cu} and _{29}^{65}\textrm{Cu} isotopes are 69.50% and 30.50% respectively.

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