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3 years ago

How much heat is required to melt 26.0 g of ice at its melting point?

Chemistry

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

Heat required to melt 26.0 g of ice at its melting point is 8.66 kJ.

Explanation:

Number of moles of water in 26 g of water: 26× $\frac{1}{18.02}$ moles

=1.44 moles

The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol.

we have relation as:

q = n × ΔH

where:

q = heat

n = moles

Δ H = enthalpy

So calculating we get,

q= 1.44*6.02 kJ

q= 8.66 kJ

We require 8.66 kJ of energy to melt 26g of ice.

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The active planning stage of change is ____________. Group of answer choices maintenance action contemplation preparation

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Answer:

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Explanation:

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3 years ago

If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream

Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values of the outflows.

C=478.11kg/h

B=5522.33kg/h

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3 years ago

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3 years ago

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