Chemistry, 18.03.2021 01:00 jordanmjdmary8571

9.32 L of pentane reacts with 36.4 L of oxygen. how many liters of carbon dioxide are formed?

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Answer from: azolber

22.75 L CO2

Explanation:

Balance the Equation

C5H12 + O2 ==> CO2 + H2O Balance the Carbons

C5H12 + O2 ==> 5CO2 + H2O Now balance the Hydrogens

C5H12 + O2 ==> 5CO2 + 6H2O Now count up the oxygens on the right.

5*O2 + 6O ==> 10 O + 6 O ===> 16 Oxygens.

You have 2 on the left. You have to multiply the oxygens by 8

C5H12 + 8O2 ==> 5CO2 + 6 H2O

The equation is now balanced.

Now find out how much oxygen is needed to burn 9.32 L of Pentane

1 Liter Pentane / 8 Liters Oxygen = 9.32 L Pentane / x Cross multiply

x = 8 * 9.32

x = 74.56

Read this next sentence very carefully. You need 74.56 Liters of oxygen to burn all the pentane you have. There's not that much there. You only have 36.4 Liters of Oxygen. Some of the pentane won't get burned. Consequently there will be a shortage of CO2 as well.

Find out the number of Liters of Pentane that will be burned up.

1 / x = 8 / 36.4 Cross multiply

8x = 36.4 Divide by 8

8x/8 = 36.4/8

x = 4.55 Liters of Pentane will be used up.

Find the number of Liters of CO2

For every Liter of Pentane, 5 Liters of CO2 will be produced.

1/5 = 4.55/x Cross Multiply

x = 5 * 4.55

x = 22.75

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.

Answer from: Quest

Answerhave membrane-bound organelles;

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