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7nadin3 [17]
3 years ago
7

5.00 g of carbon were adiabatically burned to CO2 in a 2.00 kg copper calorimeter which contained 2.50 kg of water. The temperat

ure increased by 14.1 °C.
Specific heats: copper, 0.385 J K⁻¹ g⁻¹; water, 4.184 J K⁻¹ g⁻¹.

a) Determine the heat of combustion of carbon ignoring any heat absorbed by the calorimeter.
b) Now determine the heat of combustion considering the heat absorbed by the calorimeter.
Chemistry
1 answer:
lorasvet [3.4K]3 years ago
3 0

Answer:

A. -163.96kJ

B. -158.34kJ

Explanation:

Heat of combustion is the heat released when an element with oxygen at stp.

Reaction for the combustion of carbon:

C(s) + O2(g) --> CO2(g)

Enthalpy heat of combustion, C (using Hess law) = -393.5 kJ/mol

m = 5g

Molecular weight = 12 g/mol

No of moles = mass/molecular weight

= 5/12

= 0.417mol

DH = -393.5 * 0.417

= -163.96 kJ

B. Heat absorbed by the calorimeter = heat evolve by combustion

= mCH2ODT + mCcDT

Where CH2O is the specific heat capacity of the water in the calorimeter

Cc is the specific heat capacity of copper calorimeter

= (2500*4.184*14.1) + (2000*0.385*14.1)

= -158343J

= -158.34kJ

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Answer:

1.9 L

Explanation:

Step 1: Given data

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Step 2: Calculate the final volume of the balloon

According to Avogadro's law, the volume of an ideal gas is directly proportional to the number of moles. We can calculate the final volume of the balloon using the following expression.

V₁ / n₁ = V₂ / n₂

V₂ = V₁ × n₂ / n₁

V₂ = 2.5 L × 3.0 mol / 4.0 mol

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aluminum has a density or 2.7g/cm3. copper has a density of 8.96/cm3. which metal would you choose to build a model airplane?
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Flow of heat energy through liquids and<br> gases
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And with solution...
Tems11 [23]

Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

N₂ = 5.6 g

Volume of NH₃ = ?

                              14 g of N   ----------------  1 mol

                              5.6 g -----------------------   x

                             x = (5.6 x 1) / 14 = 0.4 mol of N

Reaction

                                N₂    +     3H₂    ⇒    2NH₃

                                1 mol of N₂   ----------------  2 moles of NH₃

                                0.4 mol of N₂ --------------   x

                               x = (0.4 x 2) / 1

                               x = 0.8 mol of NH₃

Formula

                 PV = nRT

P = 5200 torr = 6.84 atm

V = ?

n = 0.8

R = 0.082 atm L/ mol °K

T = 450°C = 723°K

Substitution

                     V = (0.8)(0.082)(723) / 6.84

                     V = 6.93 L

7 0
3 years ago
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