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Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutr

Solnce55 [7]

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

$Mg(OH)_{2}\Leftrightarrow Mg^{2+}(aq)+2(OH)^{-}(aq)$

$[Mg^{2+}][OH^{-}]$

Initial 0 0

Equili +S +2S

Final S 2S

$K_{sp}=[Mg^{2+}][OH^{-}]$

$5.61\times 10^{-11}=(S)(2S)^{2}$

$S=(\frac{5.61\times 10^{-11}}{4})^{1/3}=2.41\times 10^{-4}M$

Solubility of $Mg(OH)_{2}$ in 0.180 M NaOH is a follows.

$Mg(OH)_{2}\Leftrightarrow Mg^{2+}(aq)+2(OH)^{-}(aq)$

$[Mg^{2+}][OH^{-}]$

Initial 0 0

Equili +S +2S

Final S 2S+0.180M

$K_{sp}=[Mg^{2+}][OH^{-}]$

$5.61\times 10^{-11}=(S)(2S+0.180)^{2}$

$S=1.46\times 10^{-9}M$

$Ratio\,of\,solubility=\frac{2.41\times 10^{-4}}{1.46\times 10^{-9}}=165068.49$

Therefore, The ration of the molar solubility is 165068.49.

4 0

3 years ago

On a mission to a newly discovered planet, an astronaut finds gallium abundances of 61.29% for 69Ga and 39.71% for 71Ga. What is

Verizon [17]

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element(68.72*61.29) + (39.71*70.92) = 4211.84 + 2816.237028.07/100= 70.28

70.28 is the atomic mass of Gallium for the location

4 0

3 years ago

How many grams of Ag2S2O3 form

Art [367]

Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgBr: 2 moles
Na₂S₂O₃: 1 mole
Ag₂S₂O₃: 1 mole
NaBr: 2 moles

The molar mass of the compounds is:

AgBr: 187.77 g/mole
Na₂S₂O₃: 158 g/mole
Ag₂S₂O₃: 327.74 g/mole
NaBr: 102.9 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

AgBr: 2 moles ×187.77 g/mole= 375.54 grams
Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
NaBr: 2 moles ×102.9 g/mole= 205.8 grams

<h3>Mass of Ag₂S₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?

$mass of Ag_{2} S_{2} O_{3} =\frac{125 grams of AgBrx327.74 grams of Ag_{2} S_{2} O_{3}}{375.54 grams of AgBr}$

<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>

Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.

Learn more about the reaction stoichiometry:

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4 0

2 years ago

In each of the following diatomic molecules, which end of the molecule is positive relative to the other end?

docker41 [41]

Answer:

A- Hydrogen Fluoride, HF

3 0

3 years ago

What is the base ionization constant expression for ammonia?

4vir4ik [10]

Explanation:

Kb: The base ionization constant

2) Ammonia (formula = NH3) is the most common weak base example used by instructors. ... This constant, Kb, is called the base ionization constant. It can be determined by experiment and each base has its own unique value. For example, ammonia's value is 1.77 x 10¯5.

7 0

3 years ago