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Tema [17]
3 years ago
11

‼️Picture Shown‼️Extra points‼️ You will loose your points if your answer is removed :) ‼️So trolls don’t waste your time on thi

s one ‼️
Use the table to calculate the Hf for the following reaction. Please show all work.

C6H6(g) + 15 O2 ——> 12 CO 2 (g) + 6 H2O(g)

Chemistry
1 answer:
insens350 [35]3 years ago
6 0

C6H6 + 15 O2 -- > 12 CO 2 + H2O

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Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

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Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has pre
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Why is it the principle about the relationship between mass and volume always true of substance, but not always true of mixtures
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Answer:

For pure substances, the mass and volume will always be the same or will always change the same way because all substance are the same throughout.

While for mixtures, you can have varying amount of each component therefore mass and volume will not change the same way for substances

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What is heat of vaporization?
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B. It is the heat required to change a gram of substance from solid to liquid

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