Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
Answer:
Element 2
Explanation:
If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.
The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.
CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4
1 mol CH4 = 16 g
16 g ---- 891,6 kJ
x g ----- 272 kJ
x = 272 kJ × 16 g / 891,6 kJ = 4,88 g
You must burn 4,88 g of CH4.
:-) ;-)
The answers
A) a
B) a
C) an
D) the
