Answer: A. The oceans‘ tidal would be smaller because the moon would exert less gravitational pull on earths oceans.
Explanation:
i got it right :)
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
Learn more about heat transfer:
brainly.com/question/10286596
#SPJ1
Its a strong electrolyte!.
Answer:
The mole fraction of codeine is 5.4%
Explanation:
Mole fraction = Mole of solute / Total moles (Mole solute + Mole solvent)
Solute (Codeine)
Molar mass 299.36 g/m
Mass / Molar mass = Mole → 46.85 g /299.36 g/m = 0.156 moles
Solvent (Ethanol)
Molar mass 46.07 g/m
125.5 g / 46.07 g/m = 2.724 moles
Mole fraction (Codeine) 0.156 / (0.156 + 2.724) → 0.054
<span> Molar mass (H2)=2*1.0=2.0 g/mol
Molar mass (F2)=2*19.0=38.0 g/mol
Molar mass (HF)=1.0+19.0=20.0 g/mol
5.00 g H2 * 1mol H2 /2 g H2=2.50 mol H2
38.0 g F2*1mol F2/38.0 g F2=1.00 mol F2
H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 1 mol
From problem 2.50 mol 1 .00mol
We can see that excess of H2, and that F2 is a limiting reactant.
So, the amount of HF is limited by the amount of F2.
</span> H2(g) + F2(g) → 2 HF(g)
From reaction 1 mol 2 mol
From problem 1.00 mol 2.00mol
2.00 mol HF can be formed.
2.00 mol HF*20.0g HF/1mol HF=40.0 g HF can be formed
