Chemistry, 13.03.2020 03:53, leeplott4747

A total of 25.6 kj of hear energy is added to a 5.18l sample of helium at .987atm The gas is allowed to expand against a fixed external pressure to a volume of 18.4l

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A) calculate the work done on or by the helium gas units of joules, J.

B) what is the change in the helium a internal energy in units kilojoules, KJ?

A. Workdone by helium, W = -1.322KJ

B. Internal energy, DE = 24.278KJ

Explanation:

Workdone can be defined as the force moving through a distance. For a gaseous system, when the volume of the gas expands, the system is losing energy. Therefore,

W = -P*DV

Where P is the pressure in pascal

DV is the change in volume in m3

DV = Vfinal - Vinitial

= 18.4 - 5.18

= 13.22L

W = -(0.987 * 13.22)

= -13.0481L.atm

Converting L.atm to joule,

= -13.0481 * 101.325

= -1322.099J

= -1.322KJ

If the system loses heat, Q the rection occurring is Exothermic.

Heat is the transfer of energy from one system to another.

Q = mcDT

Where m is the mass of the system

C is the specific heat capacity

Q = 25.6KJ

Internal energy is the summation of the heat supplied to a system and the workdone by the system

DE = Q + W

DE = 25.6 + (-1.322)

= 24.278KJ

answered: Guest

Idon’t know what the answer is i wish i could

answered: Guest

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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A total of 25.6 kj of hear energy is added to a 5.18l sample of helium at .987atm The gas is allowed...