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Y_Kistochka [10]
3 years ago
14

The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate

freezes at 84.0 C. Be sure your answer contains a unit symbol and the correct number of significant digits
Chemistry
1 answer:
Hitman42 [59]3 years ago
7 0

<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

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_HNO3 + _Al(OH)3
bekas [8.4K]

Answer:

Option D. 3, 1, 3, 1

Explanation:

From the question given above,

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

The equation can be balance as follow:

HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:

3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃

There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:

3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃

Now, the equation is balanced.

Thus, the coefficients are 3, 1, 3, 1

7 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

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

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

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:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

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azamat

Answer:

You can view more details on each measurement unit: molecular weight of Copper(I) Oxide or grams The molecular formula for Copper(I) Oxide is Cu2O. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Copper(I) Oxide, or 143.0914 grams.

Explanation:

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