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Kaylis [27]
3 years ago
15

Hydrogenotrophy is Choose one: A. an award given to fuel companies that have the most fuel-efficient vehicles. B. the oxidation

of water during photosynthesis to liberate electrons, protons, and oxygen gas. C. the use of hydrogen gas as an electron donor. D. the generation of hydrogen gas by methanogens in syntrophy.
Chemistry
1 answer:
Mrrafil [7]3 years ago
8 0

Answer:

C. the use of hydrogen gas as an electron donor.

Explanation:

Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.

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How many g of Fe2O3 will be produced from 132.7 g of o2
zhuklara [117]

Answer:

73.46839716589713698731965

Explanation:

4 0
3 years ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

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3 years ago
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Elements K and L have 12 and 17 protons respectively. If k has 24 neutrons and L has 36, in a tabular form identify the number o
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Arrange the elements in order of increasing ionization energy. Use the
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Answer:

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