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muminat
4 years ago
14

An ideal gas in a sealed container has an initial volume of 2.65 L. At constant pressure, it is cooled to 25.00 ∘C, where its fi

nal volume is 1.75 L. What was the initial temperature?
Chemistry
1 answer:
svet-max [94.6K]4 years ago
8 0

Answer:

Initial temperature was: 178.3^\circ \:C

Explanation:

The Relation between the volume and temperature at constant pressure is:

\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{2.65}{T_1}=\frac{1.75}{\left(25+273\right)}\\\\T_1=451.3\:K\quad\:or\:\quad 178.3^\circ\:C

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What Size container do you need to hold 0.0459 mol of N2 gas at stp
Allushta [10]
Hello!

You need to calculate the volume of the container. To calculate the volume of this amount of N₂ gas we need to make the assumption that N₂ behaves like an ideal gas. 

1 mole of an ideal gas under Standard Temperature and Pressure occupies 22,4 L so the calculations are as follows:

0,0459 mol N_2* \frac{22,4 L}{1 mol N_2}= 1,028 L

So, the volume of the container should be 1,028 L or more.

Have a nice day!
4 0
3 years ago
Suppose that CO2 in the atmosphere was held at a constant, natural level for a few thousand years. Then, CO2 was added to double
MissTica

Answer:

temperature before the increase in CO2 was a few degrees lower than temperature after the increase.

Explanation:

CO2 in the atmosphere has the property of trapping heat by absorbing it. So, with increase in the level of CO2 in the atmosphere the more heat will be absorbed by it and hence the temperature before the increase in CO2 was a few degrees lower than temperature after the increase.

Level of CO2 ∝ Temperature of the Earth

8 0
3 years ago
In a chemical process, three bottles of a standard fluid are emptied into a larger container. A study of the individual bottles
Setler79 [48]

Answer:

a

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b

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Explanation:

The explanation of this answer is shown on the first uploaded image

5 0
3 years ago
What's a member of the boron family and the most abundant metal in the earths crust
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4 0
4 years ago
The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
ValentinkaMS [17]

Answer:

"6.7\times 10^{-4} \ atm" is the right answer.

Explanation:

Given:

Partial pressure of N_2,

= 0.20 atm

Partial pressure of H_2,

= 0.15 atm

K_p = 1.5\times 10^3 at 400^{\circ} C

As we know,

⇒ K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}

By putting the values, we get

    1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}

        pNH_3^2 = \frac{0.000675}{1.5\times 10^3}

                    =6.7\times 10^{-4} \ atm

                   

3 0
3 years ago
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