Chemistry, 17.09.2019 10:30 autumnplunkett09

The volume of helium in a blimp is 6.28 × 109 milliliters. the density of helium in the blimp is 0.1786 . find the mass of the helium in the blimp. (hint: 1,000 l = 1 cubic meter.)
a.
1,120 kg
b.
1.12 kg
c.
3.52 × 107 kg
d.
2,840 kg

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Answers

The volume of helium in a blimp is 6.28 \times 10^{9} millilitres. The density of helium in the blimp is 0.1786. Then the mass of the helium in the blimp is 1,120 kg

Option A

Explanation:

The ratio of amount of matter (mass) an object has and its volume is defined as density. If an object has more amount of matter, then it has more density. The formula for density is,

$\text { Density }=\frac{\text { mass }}{\text { volume }}$

Given data,

In blimp, helium’s volume = $6.28 \times 10^{9} \mathrm{ml}$

Helium’s density = $0.1786 \frac{k g}{m^{3}}$

As per conversion factor, we know that

$1 \mathrm{mm}^{3}=10^{-9} \mathrm{m}^{3}$

So, Helium’s volume will be $6.28 m^{3}$

Rearranging the formula,

$\text { mass }=\text { Density } \times \text { volume }$

$=0.1786 \frac{\mathrm{kg}}{\mathrm{m}^{3}} \times 6.28 \mathrm{m}^{3}$

$= 1120 \mathrm{kg} \text { is the Answer }$

Answer from: Quest

the answer is a. co2, this would be colder than the : )

explanation:

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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