Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
6.8x10^5? Scientific notation
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
