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2 years ago

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500 mL solution of calcium chloride with a concentration of 0.75 mol/L. The mass of solute needed to make this solution is ___ _

__ ___ ___ g.

1 answer:

BabaBlast [244]2 years ago

5 0

Answer:

41.63g

Explanation:

Given parameters:

Volume of CaCl₂ = 500mL = 0.5L

Concentration = 0.75mol/L

Unknown:

Mass of the solute needed = ?

Solution:

The mass of the solute can be derived using the expression below;

Mass = number of moles x molar mass

But,

Number of moles = Concentration x Volume

So;

Mass = Concentration x Volume x molar mas

Molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol

Mass = 0.75 x 0.5 x 111 = 41.63g

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The boiling point of bromine is 59 °C. Which of the following best predicts the boiling point of iodine monochloride, a polar co

soldier1979 [14.2K]

Answer:

Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.

Explanation:

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3 years ago

Read 2 more answers

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

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3 years ago

Stomach acid is approximately 0.10 M HCl. How many mL of stomach acid can be neutralized by one regular antacid tablet that cont

Radda [10]

Answer:

100 mL

Explanation:

The reaction that takes place is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

First we <u>convert 500 mg of CaCO₃ into mmoles</u>, using its <em>molar mass</em>:

500 mg ÷ 100 mg/mmol = 5 mmol CaCO₃

Then we <u>convert 5 mmoles of CaCO₃ into HCl mmoles</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

5 mmol CaCO₃ * $\frac{2mmolHCl}{1mmolCaCO_3}$ = 10 mmol HCl

Finally we <u>calculate the volume of a 0.10 M HCl solution (such as stomach acid) that would contain 10 mmoles</u>:

10 mmol / 0.10 M = 100 mL

7 0

2 years ago

Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/

ankoles [38]

The balanced chemical equation for the combustion of butane is:

$2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)$

Δ $H_{reaction}^{0}$ = Σ $n_{products}$ Δ $H_{f}^{0}_{(products)}$ -Σ $n_{reactants}$ Δ $H_{f}^{0}_{(reactants)}$

= $[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]$ =[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

= -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

$1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}$

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

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2 years ago

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Dmitrij [34]

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3 years ago