Answer:
91.7 kJ
Explanation:
Step 1: Given data
- Mass of ammonia (m): 66.7 g
- Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol
Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia
The molar mass of ammonia is 17.03 g/mol.
66.7 g × 1 mol/17.03 g = 3.92 mol
Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia
We will use the following expression.
Q = ΔH°vap × n
Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ
Eventually it would leak out creating a rather bland soda pop!
Have a nice day!
:D
Answer:
7.54%
Explanation:
A hydrate is a salt that has molecules of water incorporated into the crystals. It is represented by the molecular formula of the salt followed by how many molecules of water it has: XY.nH₂O.
So, the mass of water in the sample will be the difference between the hydrate and the salt without water:
mass of water = 1.034 - 0.956 = 0.078 g
The mass percentage is the mass of water divided by the total mass, and then multiplied by 100%:
(0.078/1.034)x100%
7.54%
Answer is: mass fo ammonium chloride is 93.625 grams.
V(NH₄Cl) = 5 L.
c(NH₄Cl) = 0.35 M.
n(NH₄Cl) = V(NH₄Cl) · c(NH₄Cl).
n(NH₄Cl) = 5 L · 0.35 mol/L.
n(NH₄Cl) = 1.75 mol.
M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.
m(NH₄Cl) = n(NH₄Cl) · M(NH₄Cl).
m(NH₄Cl) = 1.75 mol · 53.5 g/mol.
m(NH₄Cl) = 93.625 g.
And I don't exactly get what you are trying to ask hit me back up though
