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saveliy_v [14]
1 year ago
10

Calculate the volume in litres (L) of 1.41 mol of gas at 68.0 kPa and 27.00 °C (gas constant R = 8.314 J mol K-.).

Chemistry
1 answer:
Pani-rosa [81]1 year ago
3 0

Answer:

3.) 51.7 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In the equation,

-----> P = pressure (kPa)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (8.314 kPa*L/mol*K)

-----> T = temperature (K)

First, you need to convert the temperature from Celsius to Kelvin. Then, you can plug the given values into the equation and simplify to find "V".

P = 68.0 kPa                          R = 8.314 kPa*L/mol*K

V = ? L                                    T = 27.00 °C + 273 = 300 K

n = 1.41 moles

PV = nRT                                                              <----- Ideal Gas Law

(68.0 kPa)V = (1.41 moles)(8.314 kPa*L/mol*K)(300 K)      <----- Insert values

(68.0 kPa)V = 3516.822                                      <----- Multiply right side

V = 51.7                                                                <----- Divide both sides by 68.0

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A student used 10mL water instead of 30mL for extraction of salt from mixture. How may this change the percentage of NaCl extrac
34kurt
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3

Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
</span>
7 0
3 years ago
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