A solution was made by adding water to 0.250 g of H2Z until the volume totaled 25.00 mL. Subsequent titration required 40.50 mL
1 answer:
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[H_{3}O^{+}] = 0.00770 M
The equilibrium equation representing the dissociation of
Given [H_{3}O^{+}] = 0.00770 M
Let the initial concentration of acid be x and change y
So y = =
= 0.00770 M
0.00257 x - 0.00001979 = 0.00005929
x = 0.031 M
Therefore, initial concentration of the weak acid is <u>0.031 M</u>
Answer:
Partial pressure of in the gas was 733 torr and mass of
in the sample was 2.12 g.
Explanation:
a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of )
Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.
So, partial pressure of = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr
b) Lets assume that behaves ideally. Hence-
PV=nRT
where P is pressure of , V is volume of
, n is number of moles of
, R is gas constant and T is temperature in kelvin
here P = 733 torr = = 0.9646 atm
V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K
So,
=
= 0.0259 moles
As 3 moles of are produced from 2 moles of
therefore 0.0259 moles of
are produced from
moles or 0.0173 moles of
.
Molar mass of = 122.55 g
So mass of in sample =
= 2.12 g