Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.
(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of 
will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of will increase then.
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Answer:
0.00268 M
Explanation:
To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.
<u>Step 1:</u>
3.00 mL / 1,000 = 0.00300 L
Molarity = moles / volume (L)
0.0250 M = moles / 0.00300 L
(0.0250 M) x (0.00300 L) = moles
7.50 x 10⁻⁵ = moles
<u>Step 2:</u>
25.0 mL / 1,000 = 0.0250 L
0.0250 L + 0.00300 L = 0.0280 L
Molarity = moles / volume (L)
Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)
Molarity = 0.00268 M
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