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umka21 [38]
1 year ago
11

you have 150.0 ml of a 0.807 m solution of ce(no₃)₄. what mass (in grams) of ce(no₃)₄ would be required to make the solution?

Chemistry
1 answer:
Hitman42 [59]1 year ago
4 0

The mass of Ce(NO3)₄ that will be needed to create the solution was 24.16 g (in grams).

<h3>What exactly is the solution?</h3>

A homogenous mixture with one or even more solutes that have been dissolved in the solvent is known as a solution. The material that a solute dissolves in to create a homogenous mixture is known as a solvent. To create a homogenous mixture, a component must dissolve in a solvent.

<h3>What is an illustration of a solution?</h3>

One nanometer or smaller particles are referred to as a solution in a homogeneous mixture made up of two or more components. There are many different types of solutions, including sugar and salt solutions and fizzy water. Each component shows up as a different phase in a solution.

<h3>Briefing:</h3>

Mole of Ce(NO₃)₄ = 0.06225 mole

Molar of Ce(NO₃)₄ = 140.12 + 4[14 + (16×3)]

= 140.12 + 4[14 + 48]

= 140.12 + 4[62]

= 140.12 + 248

= 388.12 g/mol

Mass of Ce(NO₃)₄ = 0.06225 × 388.12

Mass of Ce(NO₃)₄ = 24.16 g

To know more about solution visit:

brainly.com/question/1616939

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Answer:

Correct option is

B

5 liters of CH

4

(g)NO

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at STP

No. of molecules=

22.4

5

mol=

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5

×N

A

molecules

A) 5ℊ of H

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No. of moles=

2

5

mol=

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×N

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B) 5l of CH

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=

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5

mol=

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5

N

A

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C) 5 mol of O

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D) 5×10

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molecules of CO

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