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labwork [276]
3 years ago
13

The temperature inside your freezer is 0 degrees Celsius. You place a balloon with an initial temperature of 30 degrees C and a

volume of 0.75 liters in the freezer. Once the balloon is fully chilled, what will its volume be?
Remember to pay close attention to the units of temperature before beginning your calculations.
Chemistry
1 answer:
blsea [12.9K]3 years ago
4 0

Answer:

V=0.68L

Explanation:

For this question we can use

V1/T1 = V2/T2

where

V1 (initial volume )= 0.75 L

T1 (initial temperature in Kelvin)= 303.15

V2( final volume)= ?

T2 (final temperature in Kelvin)= 273.15

Now we must rearrange the equation to make V2 the subject

V2= (V1/T1) ×T2

V2=(0.75/303.15) ×273.15

V2=0.67577931717

V2= 0.68L

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1 lb of CO2 occupies 0.6 ft^3 at a pressure of 200 psi. Determine the temperature of the system.
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<u>Answer:</u> The temperature of the system is 273 K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

Given mass of carbon dioxide = 1 lb = 453.6 g   (Conversion factor: 1 lb = 453.6 g)

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

\text{Moles of carbon dioxide}=\frac{453.6g}{44g/mol}=10.31mol

To calculate the temperature of gas, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide = 200 psia = 13.6 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of carbon dioxide = 0.6ft^3=16.992L    (Conversion factor:  1ft^3=28.32L )

n = number of moles of carbon dioxide = 10.31 mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the system = ?

Putting values in above equation, we get:

13.6atm\times 16.992L=10.31mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=273K

Hence, the temperature of the system is 273 K

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Explanation:

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The products of this reaction between aluminum and sulfuric acid are two: hydrogen and aluminum sulfate.

<h3>What are the products in a reaction?</h3>

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