Answer:
The answer is 20 % V/V
Explanation:
We use this formula for calculate the %V/V:
%V/V= (ml solute/ml solution) x 100= (75ml/375 ml)x 100 = 20 % V/V
<em>The% V / V represents the amount of ml of solute dissolved in 100 ml of solution</em>
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Decreasing the temperature of the reaction,the reaction shifts forward.
The explanation is given below.
Explanation:
If the temperature of the reaction mixture is increased,then the equilibrium will shift to decrease the temperature.
If the temperature of the reaction mixture is decreased,then the equilibrium will shift to increase the temperature.
During the formation of the ammonia,it gives off heat.So it is an exothermic reaction.
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A decrease in the temperature favors the reaction that is exothermic (the forward reaction)because it produces energy.Therefore,if the temperature is decreased,the yield of the ammonia increases.
<em>Therefore if the temperature is increased,the reaction shifts forward and the yield of the ammonia increases and it is an exothermic reaction.</em>
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
Answer:
3Mg(NO3)2(aq)+2Na3PO4(aq)⇒Mg3(PO4)2(s)+6NaNO3(aq)
Explanation:
