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TEA [102]
3 years ago
7

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

ollowing equations:
(a) HNO3+H2O⟶H3O++NO3−
(b) CN− +H2O ⟶ HCN+OH−
(c) H2SO4+Cl− ⟶HCl+HSO4−
(d) HSO4−+OH− ⟶SO42−+H2O
(e) O2− + H2 O ⟶ 2OH−
Chemistry
1 answer:
julia-pushkina [17]3 years ago
6 0

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-

HNO_3 is Bronsted lowry acid and NO_3^- is its conjugate base.

H_2O is Bronsted lowry base and H_3O^+ is its conjugate acid.

(b)

CN^- + H_2O \rightarrow HCN + OH^-

CN^- is Bronsted lowry base and HCN is its conjugate acid.

H_2O is Bronsted lowry acid and OH^- is its conjugate base.

(c)

H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-

H_2SO_4 is Bronsted lowry acid and HSO_4^- is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O

HSO_4^- is Bronsted lowry acid and SO_4^{2-} is its conjugate base.

OH^- is Bronsted lowry base and H_2O is its conjugate acid.

(e)

O_{2-}+H_2O \rightarrow 2OH^-

O_{2-} is Bronsted lowry base and OH- is its conjugate acid.

H_2O is Bronsted lowry acid and OH- is its conjugate base.

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Explanation:

We are given:

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According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

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2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

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Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

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