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11 months ago

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]11 months 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

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How do I calculate the molecular weight of Mg(OH)2?

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<h3>Further explanation</h3>

Given

Mg(OH)2 compound

Required

The molecular weight

Solution

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Relative molecular weight (M) : The sum of the relative atomic mass of Ar

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2 years ago

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

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Explanation :

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$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

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The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

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Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

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Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

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