What is SnO2 compound name
1 answer:
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Explanation:
The given data is as follows.
[HCOOH] = 0.2 M, [NaOH] = 2.0 M,
V = 500 ml, [Benzoic acid] = 0.2 M
First, we will calculate the number of moles of benzoic acid as follows.
No. of moles of benzoic acid = Molarity × Volume
=
= 0.095 mol
And, moles of NaOH present in the solution will be as follows.
No. of moles of NaOH = Molarity × Volume
=
= 0.05 mol
Hence, the ICE table for the chemical equation will be as follows.
Initial: 0.095 0.05 0 0
Equlbm: (0.095 - 0.05) 0 0.05
pH =
=
= 4.245
For,
Initial: 0.2x 2(0.5 - x) 0
Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)
As,
pH =
4.245 = 3.75 +
= 0.5
= 3.162
Now,
= 3.162
x = 0.464 L
Volume of NaOH = (0.5 - 0.464) L
= 0.036 L
= 36 ml (as 1 L = 1000 mL)
And, volume of formic acid is 464 mL.
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong