Hydrogen reacts with oxygen based on the following equation:
2 H2<span> + O</span>2<span> → 2 H</span>2<span>O
</span>
From the periodic table:
molar mass of hydrogen = 1 gram
molar mass of oxygen = 16 grams
From the balanced equation above, we can find that:
4 grams of hydrogen react with 32 grams of oxygen to produce 36 grams of water.
This means that: 0.73 grams of hydrogen require (0.73x32) / 4 = 5.84 grams of oxygen to react with.
Since only 3.28 grams of oxygen are reacting, this means that oxygen is our limiting reagent and that the reaction would stop once the amount of oxygen is consumed.
So, we will base our calculations on oxygen.
mass of water produced from 3.28 grams of oxygen can be calculated as follows:
mass of water = (3.28 x 36) / 32 = 3.69 grams
Explanation:
u can put them in order crossed are spectator ions. hope this helps:)
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
B) It will accelerate to the right because 500 N> 300 N
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
