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pashok25 [27]
3 years ago
8

The ratio of the specific heat of aluminum to the specific heat of iron is 2:1. How much energy must be transferred to the alumi

num pan compared with the amount of
energy transferred to the iron pan?
Chemistry
1 answer:
erastova [34]3 years ago
3 0

Given what we know about ratios, we can confirm that in comparison to the iron pan, the aluminum pan will receive twice as much heat.

<h3>Why would the aluminum pan receive double the heat?</h3>
  • This has to do with the ratio being presented.
  • A ratio of 2:1 means that for every unit of heat to the iron, the aluminum receives 2.
  • In other words, the aluminum pan receives twice as much heat when compared to the iron pan.

Therefore, we can confirm that because the ratio of heat to the aluminum in comparison to the iron pan is 2:1, this means that for every 1 unit of heat to the iron pan, the aluminum pan will receive 2. This results in double the total heat received by the aluminum pan.

To learn more about ratios visit:

brainly.com/question/1504221?referrer=searchResults

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Answer:

The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky

Explanation:

we have the following balanced reaction:

CaC2 + 2H2O = C2H2 + Ca(OH)2

the value of K for this reaction will be equal to:

K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)

if we multiply the reaction by the value of 6, we have:

6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2

Again, the value of K for this reaction will be equal to:

K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6

For the second reaction:

6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H

The value of K for this reaction:

K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)

we also have:

K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)

Thus:

K1 = K^6 * Ky

4 0
3 years ago
For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
BlackZzzverrR [31]

Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

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Explanation:

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