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nignag [31]
2 years ago
14

What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the

following equation? CHale) +20269) CO2 + 2H2008)
Chemistry
1 answer:
fomenos2 years ago
3 0

Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →       2H₂O  + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

                         CH₄              :              H₂O

                           1                 :                2

                        0.98             :            2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

V = 43.95 L

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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ
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<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules

<u>Explanation:</u>

We are given:

E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

NH_4 will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

<u>Oxidation half reaction:</u>  NH_4\rightarrow NO_2^-+6e^-     ( × 4)

<u>Reduction half reaction:</u>  O_2+4e^-\rightarrow 2H_2O   ( × 6)

<u>Overall reaction:</u> 4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O

We are given:

Molecules of NH_4 = 28

By Stoichiometry of the reaction:

4 molecules of NH_4 reacts with 6 molecules of oxygen gas

So, 28 molecules of NH_4 will react with = \frac{6}{4}\times 28=42 molecules of oxygen gas

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What is the ph of a solution that contains 1.0 l of 0.10 m ch3cooh and 0.080 m nach3coo after 0.03 moles of naoh added?
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Answer: pH = 4.996

Explanation:

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n(CH3COOH) = 0.1mol

Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.

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Using the Hendersom-Hasselbach equation,

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From literature, pKa of Ch3COOH is 4.8

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2 years ago
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