I guess it’s potential energy.
Explanation:
Number 1
Q = 4280J
mass = x
∆∅ = 34.5 - 22
∆∅ = 12.5°C
Specific heat capacity of water = 4200J
Q = mc∆∅
4280 = x(4200)(12.5)
4280 = 52500x
x = 4280/52500
x = 0.0815kg
Number 2
c = x
Q = 3750
m = 315g
∆∅ = 78 - 28.4
∆∅ = 49.6°C
Q = mc∆∅
3750 = 315×x×49.6
3750 = 15624x
x = 3750/15624
x = 0.2400J(g°C)-¹
Number 3
c = 0.448
Q = 35000
m = 454g
∆∅ = x
Q = mc∆∅
35000 = 454(0.448)(x)
35000 = 203.392x
x = 35000/203.392
x = 172.0815
∅2 - ∅1 = ∆∅
∅2 = ∆∅ + ∅1
∅2 = 172.0815 + 24
∅2 = 196.0815
∅2 = 196°C
Number 4
Q = 8750J
m = 75g = 0.075
c = 4200J/kg/°C
∆∅ = x
Q = mc∆∅
8750 = 0.075(4200)∆∅
8750 = 315×x
x = 8750/315
x = 27.7778
∅2 - ∅1 = 27.7778
∅2 = 27.7778 + ∅1
∅2 = 27.7778 + 23
∅2 = 50.7778
∅2 = 50.8°C
Number 5
Q = 38800J
c = 0.448J/(g°C)
∆∅ = 178 - 24.5
∆∅ = 153.5°C
m = x
Q = mc∆∅
38800 = x × 0.448 × 153.5
38800 = 68.768x
x = 38800/68.768
x = 564.2159
x = 564g
Answer:
Sound waves and light waves need to change places.
Explanation:
Answer:
0.440 moles of NH₃ are produced
Explanation:
First of all, we need to determine the limiting reactant by the stoichiometry.
Equation reaction is: N₂(g) + 3H₂(g) ⟶ 2 NH₃(g)
1 mol of nitrogen needs 3 moles of hydrogen to react
Therefore 0.220 moles of N₂ will need (0.220 . 3) / 1 = 0.660 moles of H₂
As we have 0.717 moles of H₂ and we need 0.660, the hydrogen is the excess reagent, therefore, the N₂ is the limiting reactant
3 moles of H₂ need 1 mol of N₂ to react
Then, 0.717 moles of H₂ will react with (0.717 . 1) / 3 = 0.239 moles of N₂
We do not have enough N₂
After complete reaction → ratio is 1:2
1 mol of N₂ reacts to produce 2 moles of ammonia
Therefore 0.220 moles of N₂ will produce (0.220 . 2) / 1 = 0.440 moles of NH₃
Seems correct. Your steps make sense, the molar ratios are right and you've done it in a clear easy to follow manner. I see no issues with it.
