Which of the following energy sources obtains its energy from gravitational potential energy?

How many molecules are there in 9.34 grams of LiCL

julia-pushkina [17]

Assuming the question refers to LiCl (Lithium chloride) which has a molecular weight 42.39. Avogadro's constant states there are 6.022 141 79x1023 molecules per mole 9.34 g LiCl is 9.34/42.39 mole (0.220 mole) LiCl 
The number of molecules is therefore 6.022 141 79x1023x 0.220 =1.326x1023 molecules

7 0

1 year ago

(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = 2.3x10⁻³M InF₃. Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = 7.9x10⁻¹⁰

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, solubility is 6.31x10⁻⁶M

3 0

2 years ago

How many ml of a 14.0 m nh3 stock solution are needed to prepare 200 ml of a 4.20 m dilute nh3 solution? hints how many ml of a

Fudgin [204]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

14 M x V1 = 4.20 M x 200 mL

V1 = 60 mL needed of the concentrated solution

5 0

2 years ago

A 5.00 gram sample of an oxide of lead PbxOy contains 4.33g of lead. Determine the simplest formula for the compound.

Valentin [98]

PbO2

You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula would be PbO2

8 0

2 years ago

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. if 4.40 mol of m

Georgia [21]

The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2

8 0

2 years ago