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Burka [1]
3 years ago
7

You have an unknownthat contains only one cation (it is oneof the twelve from this experiment).Your solution is colorless and od

orless. Treatment with HCl producesawhite precipitate.Treatment with H2SO4 produces a white precipitate. Treatment with NaOHproducesa white precipitate that dissolves when excess NaOHis added.What isthe identity of your cation
Chemistry
1 answer:
algol [13]3 years ago
5 0

Answer:

Your cation is Pb2+

Explanation:

This is the explanation by chemical reactions

HCl (l) ----> H+(aq)  +  Cl-(aq)

Pb2+(aq)  +  2Cl-(aq)  --->  PbCl2 (s) ↓

H2SO4 (l) ----> 2H+ (aq)  +  SO4-2(aq)

Pb2+(aq)  +  SO4-2(aq) ---> PbSO4 (s) ↓

NaOH (l) ---> Na+(aq) + OH-(aq)

Pb2+(aq)  +  2OH-(aq) ---> Pb(OH)2 (s) ↓

If the reaction takes place in a strong alkaline medium, lead hydroxide dissolves in excess of base

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
UNO [17]

Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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Explanation:

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