Answer: Thus, the number of moles of aluminum chloride could be produced from 27.0 g of aluminum is 1.0 mol.
Explanation:
Answer:
30.8994048km
Explanation:
19.2 mi * 1.609344 km/1mi = 30.8994048 km
A common question isHow many mile in 19.2 kilometer?And the answer is 11.930326891 mi in 19.2 km. Likewise the question how many kilometer in 19.2 mile has the answer of 30.8994048 km in 19.2 mi.
Answer:
91.41 g of LiClO₃.
Explanation:
We'll begin by calculating the number of mole of O₂ that occupied 33.8 L. This can be obtained as follow:
22.4 L = 1 mole of O₂
Therefore,
33.8 L = 33.8 L × 1 mole / 22.4 L
33.8 L = 1.51 mole of O₂
Next, the balanced equation for the reaction.
2LiCl + 3O₂ —> 2LiClO₃
From the balanced equation above,
3 moles of O₂ reacted to produce 2 moles of LiClO₃.
Therefore, 1.51 mole of O₂ will react to produce = (1.51 × 2)/3 = 1.01 mole of LiClO₃.
Finally, we shall determine the mass of 1.01 mole of LiClO₃. This can be obtained as follow:
Mole of LiClO₃ = 1.01 mole
Molar mass of LiClO₃ = 7 + 35.5 + (3×16)
= 7 + 35.5 + 48
= 90.5 g/mol
Mass of LiClO₃ =?
Mass = mole × molar mass
Mass of LiClO₃ = 1.01 × 90.5
Mass of LiClO₃ = 91.41 g
Thus, 91.41 g of LiClO₃ were obtained from the reaction.
The answer is A, mass and volume.
The equation for density, <span>ρ,</span> is ρ = m/V. m is mass and V is volume.
Answer:
Pressure = 4.81atm
Explanation:
Pressure = ?
Temperature = 20°C = (20 + 273.15)K = 293.15K
Volume = 2.50L
R = 0.082J/mol.K
n = 0.5mol
From ideal gas equation,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles of the gas
R = ideal gas constant and may varies due to unit of pressure and volume
T = temperature of the ideal gas
PV = nRT
Solve for P,
P = nRT/ V
P = (0.5 * 0.082 * 293.15) / 2.50
P = 12.01915 / 2.50
P = 4.807atm
P = 4.81atm
The pressure of the ideal gas is 4.81atm
