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valentina_108 [34]
2 years ago
7

What is the molarity of a solution made by dissolving 8.60 g of a solid with a

Chemistry
1 answer:
Dima020 [189]2 years ago
8 0

Answer:

1.43 M

Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

Mass of solid = 8.60 g

Molar mass of solid = 21.50 g/mol

Mole of solid =?

Mole = mass / molar mass

Mole of solid = 8.60 / 21.50

Mole of solid = 0.4 mole

Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

280 mL = 280 mL × 1 L / 1000 mL

280 mL = 0.28 L

Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

Mole of solid = 0.4 mole

Volume = 0.28 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.4 / 0.28

Molarity = 1.43 M

Thus, the molarity of the solution is 1.43 M.

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Please find attached the reaction of glucose with NADPH to produce sorbitol

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Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

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Thus, we can write the complete ionic equation for the reaction as shown below:

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Why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
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A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi
vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s)   ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s)   ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

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