Answer:
sorry i do not know
Explanation:
i do not know the answer you are asking sorry i was never really good in maths anyway
Answer: 600 kJ
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Explanation:
C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)
Δ⁰Hf kJ/mol -104 0 -393.5 -285.8
Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol
Δ⁰Hcomb = 2219.70 kJ/mol
n= m /MW MW c₃H₈ = 44.1 g/mol
n= 12 g/44.1 g/mol = 0.27 mol
then for 12 g the heat released will be
0.27 mol x 2219.70 kJ/mol = 600 KJ
The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.
<h3>How to calculate volume?</h3>
The volume of a gas can be calculated using the following formula:
PV = nRT
- P = pressure
- V = volume
- n = number of moles
- R = gas law constant
- T = temperature
0.75 × 0.652 = n × 0.0821 × 313
0.489 = 25.69n
n = 0.489/25.69
n = 0.019moles
Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.
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Answer:
The behavior of molecules in different phases of matter represents a balance between the kinetic energies of the molecules and the attractive forces between them. All molecules are attracted to each other. The molecules are in the solid-state. At higher temperatures, the kinetic energy of the molecules is higher.
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
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