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Aleksandr-060686 [28]
3 years ago
9

Calculate the solubility of zn(oh)2(s) in 2.0 m naoh solution. (hint: you must take into account the formation of zn(oh)2−4, whi

ch has a kf=2×1015.)
Chemistry
1 answer:
Brilliant_brown [7]3 years ago
4 0
When we have:

Zn(OH)2 → Zn2+ 2OH-  with Ksp = 3 x 10 ^-16

and:

Zn2+ + 4OH- → Zn(OH)4 2-  with Kf = 2 x 10^15
 
by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

when we assume that the solubility is X

and when K = [Zn(OH)4 2-] / [OH-]^2

        0.6 = X / (2-2X)^2    by solving this equation for X

∴ X = 0.53 m

∴ the solubility of Zn(OH)2 = 0.53 M
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If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
3 years ago
Read 2 more answers
What type of elements generally bond covalently, and how do they satisfy the octet rule?
Alex Ar [27]

Answer:

Covalent, or, elements on the right side of the ladder (on the periodic table). Covalent compounds satisfy the octet rule by sharing electrons.

3 0
3 years ago
|
lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

n_1= n_2  (1)

and we also know that:

n = M x V_2

If we replace this expression in (1) we have:

M_1 x V_1= M_2 x V_2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M_2:

4.93 x 210 =  620 xM_2

M_2 = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

#SPJ1

6 0
2 years ago
What are alternative periodic tables
aksik [14]
Tabulations of chemical elements differing in their organization from the traditional seen periodic system
5 0
3 years ago
Solid chromium (III) reacts with oxygen gas to form solid Cr2O3. What is this type of reaction?
mafiozo [28]

Answer:

.081 g of O2

Explanation:

4Cr + 3O2 -----> 2Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3

.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2

5 0
3 years ago
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