Answer:
83.8%
Explanation:
The balanced reaction equation is;
2Al(s) + 3Cl2(g) → 2AlCl3(s)
Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3
Amount of Al = 3.11g/27 g/mol = 0.115 moles
If 2 moles of Al yields 2 moles of AlCl3
Then 0.115 moles of Al yields 0.115 moles of AlCl3
For Cl2
Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles
If 3 moles of Cl2 yields 2 moles of AlCl3
0.075 moles of Cl2 yields 0.075 * 2/3 = 0.05 moles of AlCl3
Hence Cl2 is the limiting reactant
Theoretical yield of AlCl3 = 0.05 moles of AlCl3 * 133g/mol = 6.65 g
%yield = actual yield /theoretical yield * 100
%yield = 5.57 g/6.65 g * 100
%yield = 83.8%
Answer: 218.75 kPa
Explanation:
Carbon dioxide is a gas with chemical formula CO2.
Original volume of CO2 (V1) = 3.50 L
Original pressure of CO2 (P1) = 125 kPa
New pressure of CO2 (P1) = ?
New Volume of CO2 (V2) = 2.0 L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
3.50L X 125 kPa = P1 x 2.0L
437.5 L•kPa = 2.0L•P1
Divide both sides by 2.0L
437.5 L•kPa/2.0L = 2.0L•P1/2.0L
218.75 kPa = P1
Thus, the new pressure of carbon dioxide would be 218.75 kPa
Calculate the heat gained by the water first.
q = mCpΔT
m = 20.0 g
Cp = 4.186 J/g°C
ΔT = T(final) - T(initial) = 15.0°C - 10.0°C = 5.0°C
q = (20.0)(4.186)(5.0) = 419 J
This is equal to the heat lost by the metal, so calculate Cp for the metal, given:
q = -419 J (negative because heat was lost)
m = 5.00 g
ΔT = 15.0°C - 100.0°C = -85.0°C (negative because the temperature decreased)
q = mCpΔT —> Solve for Cp —> Cp = q/mΔT
Cp = -419 / (5.00 • -85.0) = 0.986 J/g°C
Answer:
20.93 g
Explanation:
From the question given above, the following data were obtained:
Heat (Q) = 3.5 KJ
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Mass (M) =?
Next, we shall convert 3.5 KJ to J. This can be obtained as follow:
1 KJ = 1000 J
Therefore,
3.5 KJ = 3.5 KJ × 1000 J / 1 KJ
3.5 KJ = 3500 J
Next, we shall determine the change in the temperature of the water. This is illustrated:
Initial temperature (T₁) = 26°C
Final temperature (T₂) = 66°C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 66 – 26
ΔT = 40 °C
Finally, we shall determine the mass of the water. This can be obtained as follow:
Heat (Q) = 3500 J
Change in temperature (ΔT) = 40 °C
Specific heat capacity (C) = 4.18 J/gºC
Mass (M) =?
Q = MCΔT
3500 = M × 4.18 × 40
3500 = M × 167.2
Divide both side by 167.2
M = 3500 / 167.2
M = 20.93 g
Therefore, the mass of the water is 20.93 g
B. Both mechanical and chemical waethering