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Vesnalui [34]
3 years ago
14

The heat capacity of chloroform (trichloromethane,CHCl3)

Chemistry
1 answer:
allsm [11]3 years ago
5 0

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}

where,

\Delta S = change in molar entropy

n = number of moles = 1.0 mol

T_f = final temperature = 300 K

T_i = initial temperature = 273 K

C_{p,m} = heat capacity of chloroform = 91.47+7.5\times 10^{-2}(T/K)

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

\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}

\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}

\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)

\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)

\Delta S=8.626+2.025

\Delta S=10.651J/K.mol

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

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We solve as follows:

Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
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Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
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Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

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The information we have:

q=1.30MJ= 1.30*10^6 J

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Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

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Answer:

Q = -33.6kcal .

Explanation:

Hello there!

In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

Q=n*\Delta _cH

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

n=3g*\frac{1mol}{30.08g}=0.1mol

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal

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