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Murljashka [212]
2 years ago
12

Methane (CH4) reacts with excess oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). What is the percent yield of c

arbon dioxide if the actual yield is 29.8 grams and the theoretical yield is 0.814 moles?
65.6%
71.6%
83.2%
91.4%
Chemistry
2 answers:
Luda [366]2 years ago
8 0

Answer:

83.2%

Explanation:

Percent yield is the ratio of actual yield to the theoretical yield.  

Percent yield = actual yield/ theoretical yield x 100

Moles of CO₂ = 0.184 mol

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ = 0.814 mol x 44 g/mol =  35.816 g

Actual yield = 29.8 g

Theoretical yield =    35.816 g

Percent yield = 29.8 g/   35.816 g x 100 = 83.2%

musickatia [10]2 years ago
4 0
(29.8 g) / [0.184 mol (44.00964 g CO2/mol)] =0.832= 83.2% yield CO2

(hope this helps)
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Answer:

So she is very anxious because she has to wait 345600 seconds

Explanation:

60 second = 1 minute

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1 hour has 3600 seconds (60*60)

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1 day has 86400 seconds so in four days

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7 0
2 years ago
Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM
7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will  B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: <u><em>The value of Kc remains the same.</em></u>

<u><em> </em></u>As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, <u><em>The reaction must: A. run in the forward direction to reestablish equilibrium</em></u>.

By decreasing the volume, and so that Kc remains constant, being:

Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2}  ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume}  } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume

 where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so,  the number of moles of Cl₂ should decrease.<u><em>The number of moles of Cl2 will  B. decrease.</em></u>

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, <em><u>the value of Qc C. is less than Kc.</u></em>

3 0
2 years ago
If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov
artcher [175]
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present

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Answer:

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=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

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