You’ll need to be sure to count all the atoms in each side of the chemical equation.
Answer:
Use the formula PV=nRT to get the temperature
Answer:
I) the heat capacity of ammonia(s)
II) the heat capacity of ammonia(ℓ)
IV) the enthalpy of fusion of ammonia
Explanation:
Initially, ammonia at 200 K is liquid. To calculate the change of enthalpy from 200 K to 195 K (melting point) we need to know the heat capacity of ammonia(ℓ).
At 195, ammonia is in the transition from liquid to solid (solidification). To calculate the change of enthalpy in that process we need to know the enthalpy of solidification of ammonia, which has the same value but opposite sign to the enthalpy of fusion of ammonia.
From 195 K to 0 K, ammonia is solid. To calculate the change of enthalpy in that process we need to know the heat capacity of ammonia(s).
Answer:
Explanation:
2CH₃(CH₂)₆CH₃ + 25O₂ = 16CO₂ + 18H₂O.
2 moles 25 moles 18 moles
mol weight of octane = 114
72 g of octane = 72 / 114 = .6316 moles
144 g of oxygen = 144 / 32 = 4.5 moles of oxygen.
2 moles of octane reacts with 25 moles of oxygen
.6316 moles of octant will react with 25 x .6316 / 2 moles of oxygen
= 7.895 moles
So oxygen is the limiting reagent
25 moles oxygen produces 18 moles water
4.5 moles of water produces 18 x 4.5 / 25 moles of water
= 3.24 moles of water
= 3.24 x 18 grams of water
= 58.32 grams of water .
