Explanation:
As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.
Hence, amount of diethyl ether present will be calculated as follows.
(100ml - 1.468 ml)
= 98.532 ml
So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.
Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.
Amount of water = 
= 0.3427 ml
Now, when magnesium dissolves in water then the reaction will be as follows.
Molar mass of Mg = 24.305 g
Molar mass of 
= 18 g
Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.
Amount of Mg = 
= 0.462 g
(19.78 x 10) + (80.22 x 11) all of them divided by 100= 10.81 amu
Answer:
9.63 L of NO
Explanation:
We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:
Mass of NH₄ClO₄ = 50 g
Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)
= 14 + 4 + 35.5 + 64
= 117.5 g/mol
Mole of NH₄ClO₄ =?
Mole = mass /molar mass
Mole of NH₄ClO₄ = 50/117.5
Mole of NH₄ClO₄ = 0.43 mole
Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:
3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the balanced equation above,
3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.
Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.
Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:
1 mole of NO = 22.4 L
Therefore,
0.43 mole of NO = 0.43 × 22.4
0.43 mole of NO = 9.63 L
Thus, 9.63 L of NO were obtained from the reaction.
Answer:
The energies of combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g; Hydrogen = 162 kJ/g
<em>Note: The question is incomplete. The complete question is given below:</em>
To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/℃. When a 1.00-g sample of methane gas burned with
<em>excess oxygen in the calorimeter, the temperature increased by 7.3℃. When a 1.00 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of combustion (per gram) for hydrogen and methane.</em>
Explanation:
From the equation of the first law of thermodynamics, ΔU = Q + W
Since there is no expansion work in the bomb calorimeter, ΔU = Q
But Q = CΔT
where C is heat capacity of the bomb calorimeter = 11.3
kJ/ºC; ΔT = temperature change
For combustion of methane gas:
Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g
Q = 83 kJ/g
For combustion of hydrogen gas:
Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g
Q = 162 kJ/g
