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kotegsom [21]
3 years ago
13

What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution

Chemistry
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

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The true statement about the balanced equations for nuclear and chemical changes is; both are balanced according to the total mass before and after the change.

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Both in chemical and nuclear changes, mass is involved and in both cases, the law of conservation of mass strictly applies.

This means that for both chemical and nuclear changes; total mass before reaction must be equal to total mass after reaction.

Hence, both reactions are balanced according to the total mass before and after the change.

Learn more: brainly.com/question/22064431

3 0
3 years ago
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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
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Skydivers jump out of planes with only the backpacks on their backs. Of course, these packs contain the parachutes that keep sky
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the force of gravity

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Answer:

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Explanation:

The molar mass of carbon is 12 g/mol

which means that:

<u>1 mole of carbon atoms has a mass of 12 grams.</u>

Since, diamond is a allotrope of carbon.

Mass of  1.00-carat diamond in grams is:

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<u> Since, 1 mole of C contains 6.022×10²³ atoms of C</u>

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12 grams contains 6.022×10²³ atoms of C

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Thus,

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