"NH4+ <----> NH3 + H+
The constant of this equilibrium is: K = Kw / Kb = 1 x 10^-14 / 1.8 x 10^-5 =5.56 x 10^-10
5.56 x 10^-10 = x^2 / 0.20-x
x = [H+] =1.1 x 10^-5 M
pH = 5.0"
Answer:
Notice that the number of atoms of
K
and
Cl
are the same on both sides, but the numbers of
O
atoms are not. There are 3
O
atoms on the the left side and 2 on the right. 3 and 2 are factors of 6, so add coefficients so that there are 6
O
atoms on both sides.
2KClO
3
(
s
)
+ heat
→
KCl(s)
+
3O
2
(
g
)
Now the
K
and
Cl
atoms are not balanced. There are 2 of each on the left and 1 of each on the right. Add a coefficient of 2 in front of
KCl
.
2KClO
3
(
s
)
+ heat
→
2KCl(s)
+
3O
2
(
g
)
The equation is now balanced with 2
K
atoms,
Hope this helps :) remember your conversions and just practice it's fairly easy:)
no because they are two diffrent things
Data Given:
Time = t = ?
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 107.86/1 = 107.86 g
Amount Deposited = W = 17.3 g
Solution:
According to Faraday's Law,
W = I t e / F
Solving for t,
t = W F / I e
Putting values,
t = (17.3 g × 96500) ÷ (10 A × 107.86 g)
t = 1547.79 s
t = 1.54 × 10³ s
