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3 years ago

Which of the following conditions will result in a reaction that is spontaneous only at high temperatures?

1 answer:

6 0

You have to use the equation ΔG=ΔH-TΔS.
With that equation you know that a reaction with a positive change in enthalpy and positive change in entropy will only be spontaneous at high temperatures since ΔG needs to be negative for a reaction to be spontaneous.

I hopet this helps. Let me know if anything is unclear.

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An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?

aniked [119]

Answer:

2.103 J/C

Explanation:

Quantity of heat = Heat Capacity * Temperature change

Heat Capacity = Quantity of heat / Temperature Change

Heat Capacity = 61/29

Heat Capacity = 2.103 J/C

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3 years ago

Is this equation balanced? nh4oh+hc2h3o2=nh4c2h3o2+h2o

Usimov [2.4K]

Answer:

It IS balanced

Explanation:

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3 years ago

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A chemist needs 0.550 mol selenium for a reaction. What mass of selenium should a chemist use?

-BARSIC- [3]

Answer:

The chemist would require to use 43.43 grams.

Explanation:

In order to solve this problem we need to know how much do 0.550 moles of selenium weigh. To do that we use selenium's molar mass and multiply it by the given number of moles:

0.550 mol * 78.96 g/mol = 43.43 g

The chemist would require to use 43.43 grams.

4 0

3 years ago

Hot water deep within earth can _______ minerals and transport them someplace else.

Vedmedyk [2.9K]

Answer:

Dissolve

Hope this helps!

Explanation:

5 0

2 years ago

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H

Taya2010 [7]

Answer:

- Empirical:

$C_3H_7$

- Molecular:

$C_6H_{14}$

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

$n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC$

Moreover, the moles of hydrogen:

$n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH$

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

$C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}$

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

$C_3H_7$

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

$C_6H_{14}$

Which is hexane.

Best regards.

6 0

3 years ago