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3 years ago

How much heat is released when 1.4 mol of hydrogen fluoride are produced? H2(g) +F2(g) → 2HF(g) +536kJ (Please show work)

Chemistry

1 answer:

polet [3.4K]3 years ago

7 0

Answer:

375.2 kJ

Explanation:

H₂(g) +F₂(g) → 2HF(g) +536kJ

The information the equation above provides lets us know that when 2 mol of hydrogen fluoride (HF) are produced, 536 kJ of energy (as heat) is produced.

We can then state a rule of three:

2 mol HF -------- 536 kJ

1.4 mol HF --------- X

And solve for X:

X = 1.4 mol * 536 kJ / 2 mol

X = 375.2 kJ

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Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

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the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

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⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

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