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2 years ago

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A solution is prepared by dissolving 117 grams of salt (NaCl) in 250 milliliters of water (H₂O). What is

1 answer:

stiv31 [10]2 years ago

8 0

Taking into account the definition of molarity, the molarity of the solution is 8 M.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Molarity of the solution in this case</h3>

In this case, you know:

number of moles= 117 grams× $\frac{1 mole}{58.45 grams}$ = 2 moles, where 58.45 $\frac{grams}{mole}$ is the molar mass of NaCl
volume= 250 mL= 0.250 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

$Molarity=\frac{2 moles}{0.250 L}$

Solving:

<u><em>Molarity= 8 </em></u> $\frac{moles}{liter}$ <u><em>= 8 M</em></u>

Finally, the molarity of the solution is 8 M.

Learn more about molarity:

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Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)

Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

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ΔG° = ΔH° - T × ΔS°

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3 years ago

Using the Lewis dot structures of magnesium and oxygen, predict the ionic formula

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Answer:

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3 years ago

Convert 1.248×1010 g to each of the following units.

Leona [35]

Answer:

a) 1.248 x 10⁷ kg

b) 1.248 x 10⁴ Mg

c) 1.248 x 10¹³ mg

d) 1.248 x 10⁴ ton

Explanation:

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1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

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