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Oksi-84 [34.3K]
2 years ago
5

How many joules are needed to heat 20.0 g of Au from 10°C to 50°C? show work

Chemistry
1 answer:
AlexFokin [52]2 years ago
3 0

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of 1 g of a given substance by 1∘C.

In your case, aluminium is said to have a specific heat of 0.90Jg∘C.

So, what does that tell you?

In order to increase the temperature of 1 g of aluminium by 1∘C, you need to provide it with 0.90 J of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by 1∘C. So if you wanted to increase the temperature of 10.0 g of aluminium by 1∘C, you'd have to provide it with

1 gram0.90 J+1 gram0.90 J+ ... +1 gram0.90 J10 times=10×0.90 J

However, you don't want to increase the temperature of the sample by 1∘C, you want to increase it by

ΔT=55∘C−22∘C=33∘C

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

1∘C10×0.90 J+1∘C10×0.90 J+ ... +

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Here, acetaldehyde is reduced to form ethane.

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Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

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Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

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Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

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