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slega [8]
3 years ago
8

`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i

s 317.15 K and 0.959 atm, respectively. Suppose I don't want the pressure to change, but I want to the volume to go down to 0.378 L. What is the temperature that I need to reach when I cool down the balloon? To what temperature (in Celsius) must the balloon be cooled to reduce its volume to 0.378 L if the pressure doesn't change (remained constant)?
Chemistry
1 answer:
Andrew [12]3 years ago
3 0

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 439 mL  = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

\frac{0.439}{317.15}=\frac{0.378}{T_2}

T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15  

So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>

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What is the most mass of (NH4)2CO3 (H=1,C=12,N=14,O=16
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Answer:

96.09 g/mol

Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

N = 14.01 g/mol

H = 1.008 g/mol

O = 16.00 g/mol

C = 12.01 g/mol

Since your compound is  

(

N

H

4

)

2

C

O

3

, you need to multiply the atomic weights by their subscripts. Therefore,

Step 2

N = 14.01 g/mol × 2 =

28.02 g/mol

H = 1.008 g/mol × (4×2) =

8.064 g/mol

 

O = 16.00 g/mol × 3 =

48.00 g/mol

C = 12.01 g/mol × 1 =

12.00 g/mol

To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

molar mass of

(

NH

4

)

2

CO

3

=

(28.02 + 8.064 + 48.00 + 12.01) g/mol

=

96.09 g/mol

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