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olganol [36]
2 years ago
7

C4H10+ 02 → _CO2 + __ H20

Chemistry
1 answer:
Svetllana [295]2 years ago
4 0

Answer:a) 2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction

b) 8 (4 moles CO2 per mole butane)

Explanation:

could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O

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3 years ago
Calculate the number of moles found in 3.045x1024 atoms of helium.<br><br> PLS HELP
Kisachek [45]

Explanation:

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2 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

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1 year ago
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Answer:

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Explanation:

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Answer:

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Explanation:

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