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2 years ago

A 7.591-9 gaseous mixture contains methane (CH4) and butane

Chemistry

1 answer:

mestny [16]2 years ago

7 0

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y (1)

Where X is the mass of methane and Y the mass of butane

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

2 moles of oxygen react per mole of methane

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

13/2 moles of oxygen react per mole of methane

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y (2)

Where 16.04 and 58.12 are molar masses of methane and butane

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

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Answer: The amount of water that can be formed is 160 moles

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We are given:

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By Stoichiometry of the reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas

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2 years ago

A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

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Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

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434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

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