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igomit [66]
3 years ago
13

The conversion of 1 mole of water at 0°c to 1 mole of ice at 0°c releases 6.01 kj. what must be true about the heat absorbed dur

ing the melting of 1 mole of ice at 0°c to 1 mole of water at 0°c? it must be less than 6.01 kj. it must be greater than 6.01 kj. it must be 6.01 kj.
Chemistry
2 answers:
Stels [109]3 years ago
7 0

Answer: it must be 6.01 kJ.

Explanation:

Latent heat of freezing is the amount of heat released to convert 1 mole of liquid to 1 mole of solid at atmospheric pressure.

H_2O(l)\rightarrow H_2O(s)

Given: latent heat of freezing is 6.01 kJ.

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to 1 mole of liquid at atmospheric pressure. If latent heat of freezing is 6.01 kJ, latent heat of fusion will be same but the heat will be absorbed.

H_2O(s)\rightarrow H_2O(l)

love history [14]3 years ago
6 0
It must be 6.01 kj
................



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3 years ago
Read 2 more answers
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
seraphim [82]

Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
3 years ago
If 21.5 mol of an ideal gas is at 9.65 atm at 10.90 ∘C, what is the volume of the gas?
mamaluj [8]
<h3>Answer:</h3>

51.93 L

<h3>Explanation:</h3>

From the question we are given the following components of an ideal gas;

Number of moles = 21.5 mol

Pressure, P = 9.65 atm

Temperature, T = 10.90°C, but K= °C + 273.15

                         =284.05 k

We are required to calculate the volume of the ideal gas.

We are going to use the ideal gas equation which is given by;

PV = nRT, where P, V, T and n are the pressure, volume, temperature and moles of the ideal gas respectively. R is the ideal gas constant, 0.082057 L.atm/mol.K

To get the volume, we rearrange the formula to get;

V = nRT ÷ P

  = (21.5 × 0.082057 × 284.05 K) ÷ 9.65 atm

  = 51.93 L

Thus, the volume of the ideal gas is 51.93 L

3 0
3 years ago
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