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Mamont248 [21]
1 year ago
14

Under constant pressure a sample of hydrogen gas initially at 88°C and 9.6 L is cooled until its final volume is 3.4 L. What is

the final temperature?
Chemistry
1 answer:
Scilla [17]1 year ago
7 0

The final temperature = -145.24K or 127°C.

According to the data given,

T1= 88°C = 361.15K

V1= 9.6 L

V2= 3.4 L

T2= ?

We know that, according to Charles's law,

T1/V1=T2/V2

T2= T1*V2/V1

T2= 127°C or -145.24K

The final temperature = -145.24K or 127°C

<h3>What does Charles law state?</h3>

According to Charles' law, when the pressure is held constant, the volume of a given amount of gas is precisely proportional to its temperature on the kelvin scale.

<h3>What connection exists between volume and temperature?</h3>
  • In layman's words, the volume of a fixed mass of gas is exactly proportional to temperature at constant pressure.
  • When a constant mass of gas is cooled, its volume decreases, and when the temperature is elevated, its volume grows.

<h3>What are the applications of Charles law?</h3>
  • A hot air balloon drifting through the air is an illustration of Charles Law in action.
  • The air within the balloon is heated by a torch, which causes the air molecules to move more quickly and disperse.
  • This causes the air inside the balloon to be less dense than the air outside, which causes the balloon to float.

To learn more about Charles law visit:

brainly.com/question/16927784

#SPJ9

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
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Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
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