Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
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Answer:
a) 1.61 mol
b) Al is limiting reactant
c) HBr is in excess
Explanation:
Given data:
Moles of Al = 3.22 mol
Moles of HBr = 4.96 mol
Moles of H₂ formed = ?
What is limiting reactant =
What is excess reactant = ?
Solution:
Chemical equation:
2Al + 2HBr → 2AlBr + H₂
Now we will compare the moles:
Al : H₂
2 : 1
3.22 : 1/2×3.22 = 1.61 mol
HBr : H₂
2 : 1
4.96 : 1/2×4.96 = 2.48 mol
The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.
Moles of H₂ :
Number of moles of H₂ = 1.61 mol
Answer: increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object.
Explanation:
edge
<h2>
Answer: 6 moles</h2>
<h3>
Explanation:</h3>
3 H₂ + N₂ → 2 NH₃
↓ ↓
4 mol 3 mol
Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia
The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.
∴ the moles of NH₃ = moles of N₂ × 2
= 3 moles × 2
= 6 moles