How many molecules of butane (C4H10) are contained in a 9.213 g sample? How many hydrogen atoms are contained in this
sample?

There are 9.6 × 10^23 atoms of hydrogen in the sample.

Now we must first obtain the number of moles that are contained in 9.213 g sample of butane.

Molar mass of butane = 4(12) + 10(1) = 48 + 10 = 58 g/mol

Number of moles of butane = 9.213 g/58 g/mol = 0.16 moles

Number of hydrogen atoms in 9.213 g sample of this sample can now be obtained by 0.16 × 6.02 × 10^23 × 10 = 9.6 × 10^23 atoms of hydrogen.

Learn more about stoichiometry: link

9.03 X molecules of butane is present in 9.213 grams of the sample.

1.5 atoms of hydrogen are present in the sample 9.123 grams of butane.

Explanation:

Data given:

mass of butane  C4H10= 9.123 grams

atomic mass of butane = 58.12 grams/mole

number of molecules =?

number of hydrogen atoms in the sample=?

number of moles is calculated as :

number of moles =

putting the values in the above equation:

number of moles of butane =

                                             = 0.15 moles

number of molecules = number of moles x 6.02 x (Avagadro number)

Putting the values in above formula:

number of molecules of Butane = 0.15 X 6.023 X

                                                       = 9.03 X molecules

Number of hydrogen atoms:

1 mole of C4H10 contains 10 atoms of H

0.15 moles of C4H10 will have x moles

=

x = 0.15 atoms of hydrogen

b) mutualism

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