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ipn [44]
3 years ago
12

3.

Chemistry
1 answer:
swat323 years ago
3 0

Answer:

14.7838

Explanation:

Given that;

A student uses a solution that contains 16 g of water to conduct an experiment and at the end of an hour ; the amount have decreased by 3.5%

The mathematically illustration for this is given as

= 16 g ×  3.5%

= 16 *\frac{3.5}{100}

= 0.56  

After an hour; the amount have decreased by 16 - 0.56

= 15.44

after 2 hours the water decreased by another 4.25%

i.e  15.44 * \frac{ 4.25}{100}

⇒ 0.6562

after two hours , the amount would have decreased by:

= 15.44 - 0.6562

= 14.7838

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Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
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Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

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Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

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[H_3O^+][OH^-]=10^{-14}

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b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

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How much heat is needed to melt 100.0 grams of ice that is already at 0°C?
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<h3>What is Specific heat capacity?</h3>

Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass.

<h3>Heat needed to melt the cube of ice</h3>

The heat is needed to melt 100.0 grams of ice that is already at 0°C is calculated as follows;

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The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

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R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

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Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

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