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Schach [20]
3 years ago
15

If the temperature of a gas decreases at constant volume which of the following would occur

Chemistry
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

I need the things after "which of the following would occur".

Explanation:

If you were to maintain a constant volume while decreasing pressure, the temperature would also have to decrease. This is Boyle's law.

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Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken;
nexus9112 [7]

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

6 0
3 years ago
PLEASE HELP ME
pishuonlain [190]

Answer:

Explanation:

Li(s)+1/2 F₂(g)→LiF(s)

In this reaction one mole of product is formed from gaseous state of reactants . So this reaction will represent the reaction for which ΔH∘rxn equals to ΔH∘f .

In other reaction one mole is not produced if we balance the reaction. The reactant must be in gaseous state .  ΔH∘f . represents heat of formation .  ΔH∘rxn represents heat of reaction .

3 0
3 years ago
Can someone pls give me the answer to this question?
Svetllana [295]

Answer:

D

Explanation:

5 0
3 years ago
A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715
andrey2020 [161]

Answer:

w_{Ar}=0.814

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\  \\n=0.41mol

Next, using the molar fraction of argon, we compute the moles of argon:

n_{Ar}=0.41mol*0.715=0.29mol

And the moles of methane:

n_{CH_4}=0.41mol-0.29mol=0.12mol

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol}  }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}}  \\\\w_{Ar}=0.814

Regards.

7 0
3 years ago
Name a common greenhouse gas with only hydrogen and oxygen
AVprozaik [17]
I'm guessing water vapour. i can't think of any other gas limited to only hydrogen and oxygen

hope it helps :)
3 0
3 years ago
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