Question

1. How many joules of heat are required to raise the temperature of 750 g of water from 11.0 oC to 19.0 oC?

2. 8750 J of heat are applied to a piece of aluminum, causing a 66.0 oC increase in its temperature. The specific heat of aluminum is 0.9025 J/g oC. What is the mass of the aluminum?


3. A 250 g sample of water with an initial temperature of 98.8 oC loses 6500 joules of heat. What is the final temperature of the water?


4. 4786 Joules of heat are transferred to a 89.0 gram sample of an unknown material, with an initial temperature of 23.0 oC. What is the specific heat of the material if the final temperature is 89.5 oC?


5. A piece of copper has a temperature of 75.6 0C. When the metal is placed in 100.0 grams of water at 19.1 0C, the temperature rises by 5.5 0C. What is the mass of the metal?


6. The combustion of methane, CH4, releases 890.4 kJ/mol. That is, when one mole of methane is burned, 890.4 kJ are given off to the surroundings. This means that the products have 890.4 kJ less than the reactants. Thus, ΔH for the reaction = - 890.4 kJ. A negative symbol for ΔH indicates an exothermic reaction.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l); ΔH = - 890.4 kJ

a) How much energy is given off when 2.00 mol of CH4 are burned?



b) How much energy is released when 22.4g of CH4 are burned?


7. What is the change in enthalpy when 9.75 g of aluminum reacts with excess ammonium nitrate (NH4NO3) according to the equation:

2Al + 3NH4NO3  3N2 + 6 H2O + Al2O3 ΔH = -2030kJ



8. How much enthalpy/heat is transferred when 0.5113 g of ammonia (NH3) reacts with excess oxygen according to the following equation:

4NH3 + 5O2  4NO + 6H2O ΔH = -905.4kJ





9. According to the following reactions, would the burning of 5.50 g of methane (CH4) or propane (C3H8) release more heat?

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ΔΗ = -2043 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔΗ = -890. kJ

Please show all work and answer ASAP thank you so much

Answer 1

1. 25080 J
2. 146.9 g
3. 92.58 °C
4. 0.808 J/g°C
5. 117.09 g
6. a. 1708.8 kJ b.1246.56 kJ
7. 368.55 kJ
8. 6.81 kJ
9. 5.50 grams of methane produces more heat than 5.5 grams of propane.

Step-by-step explanation:

1. The specific heat capacity of water=4.18 J/gK

The enthalpy change is calculated using the formula: ΔH=MC∅ where ΔH is the change in enthalpy, M the mass of the substance, C the specific heat capacity of the substance and ∅ the temperature change.

Thus, ΔH= 750g × 4.18 J/gK × (19-11)K

=25080 J

2. Enthalpy change= mass of substance × specific heat capacity of the substance× Change in temperature.

ΔH= MC∅

M= ΔH/(C∅)

Substituting for the values in the question.

M=8750 J/(0.9025/g°C×66.0 °C)

=146.9 grams

3. Enthalpy change =mass × specific heat capacity × Temperature

ΔH= MC∅

∅ = ΔH/(MC)

=6500 J/(250 g × 4.18 J/g°C)

=6.22° C

Final temperature =98.8 °C - 6.22°C

=92.58 °C

4. Specific heat capacity =mass × specific heat capacity × Temperature change.

ΔH=MC∅

C= ΔH/(M∅)

Substituting with the values in the question.

C = 4786 J/(89.0 g×(89.5° C-23°C))

=0.808 J/g°C

5. Heat lost lost copper is equal to the heat gained by water.

ΔH(copper)= ΔH(water)

MC∅(copper)=MC∅(water)

M×0.385 J/g°C× (75.6°C- (19.1 °C+5.5°C))=100.0g×4.18 J/g°C×5.5 °C

M=(100.0g×4.18J/g°C×5.5°C)/(0.385 J/g°C×51 °C)

=117.09 grams.

6 (a). From the equation 1 mole of methane gives out 890.4 kJ

There fore 2 moles give:

(2×890.4)/1= 1780.8 kJ

(b) 22.4 g of methane.

Number of moles= mass/ RFM

RFM=12 + 4×1

=16

No. of moles =22.4 g/16g/mol

=1.4 moles

Therefore 1.4 moles produce:

1.4 moles × 890.4 kJ/mol=

=1246.56 kJ

7. From the equation, 2 moles of aluminium react with ammonium nitrate to produce 2030 kJ

Number of moles = mass/RAM

Therefore 9.75 grams = (9.75/26.982) moles of aluminium.

=0.3613 moles.

If 2 moles produce 2030 kJ, then 0.3613 moles produce:

(0.3631 moles×2030 kJ)/2

=368.55 kJ

8. From the equation, 4 moles of ammonia react with excess oxygen to produce 905.4 kJ of energy.

Number of moles= mass/molar mass

RMM= 14+3×1= 17

Therefore 0.5113 grams of ammonia = (0.5113 g/17g/mole) moles

= 0.0301 moles

If 4 moles produce 905.4 kJ, then 0.0301 moles produce:

(0.0301 moles×905.4 kJ)/4 moles

=6.81 kJ

9. From the equations, one mole of methane produces 890 kJ of energy while one mole of propane produces 2043 kJ.

Lets change 5.5 grams into moles of either alkane.

Number of moles= Mass/RMM

For propane, number of moles= 5.5g/ 44.097g/mol

=0.125 moles

For methane number of moles =5.5 g/ 16g/mol

=0.344 moles

0.125 moles of propane produce:

0.125 moles×2043 kJ/mol

=255.375kJ

0.344 moles of methane produce:

0.344 moles× 890 kJ/mol

= 306.16kJ

Therefore, 5.5 grams of methane produces more heat than 5.5 grams of propane.