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irina1246 [14]
3 years ago
9

When 9.36 g sodium hydroxide is dissolved in enough water to make a total volume of 965 mL, what is the concentration of the res

ulting solution?
Chemistry
2 answers:
VMariaS [17]3 years ago
8 0

Answer:

The concentration of the solution is 0.242 M

Explanation:

Step 1: Data given

Mass NaOH = 9.36 grams

Volume water = 965 mL = 0.965 L

Molar mass NaOH = 40 g/mol

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 9.36 grams / 40.0 g/mol

Moles NaOH = 0.234 moles

Step 3: Calculate concentration

Concentration = moles / volume

Concentration = 0.234 moles / 0.965 L

Concentration = 0.242 M

The concentration of the solution is 0.242 M

SSSSS [86.1K]3 years ago
5 0

Answer:

0.2425 M

Explanation:

Given mass = 9.36 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{9.36\ g}{39.997\ g/mol}

Moles\ of\ NaOH= 0.234\ mol

Given that:- Volume = 965 mL = 0.965 L ( 1 mL = 0.001 L )

So,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.234\ mol}{0.965\ L}=0.2425\ M

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2994 kJ

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2 years ago
The vapor pressure of a liquid in a closed container depends
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Answer:

(c) temperature

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3 years ago
90 POINTSSSS!!!
Katarina [22]

Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

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\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

(b)Total enthalpies of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

ΔᵣH° is negative, so the reaction is exothermic.

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