Chemistry, 11.03.2020 08:43 mdot8890

7. A certain mass of water was heated with 41,840 Joules, raising its
temperature from 22.0 °C to 28.5 °C. Find the mass of water.

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Answer from: wwesuplexcity28

mass of water = 1540 g

Explanation:

Given data:

Heat absorbed = 41840 j

Initial temperature = 22.0°C

Final temperature = 28.5 °C

Mass of water = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 28.5 °C - 22 °C

ΔT = 6.5°C

Now we will put the values in formula:

41840 J = m ×4.18 j/g.°C× 6.5°C

41840 J = m × 27.17 j/g

m = 41840 J /27.17 j/g

m = 1540 g

Answer from: Quest

The answer is nitrogen-containing bases

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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7. A certain mass of water was heated with 41,840 Joules, raising its
temperature from 22.0 °C...

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