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melomori [17]
3 years ago
14

Plz help if you can!!!

Chemistry
2 answers:
ivann1987 [24]3 years ago
7 0

Answer: the answer is d

Explanation:because atoms are what make things up and that means there are multiple molecules in one atom

professor190 [17]3 years ago
4 0
Molecules breaking down into atoms
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Abigail obtained 36.6 grams of calcium carbonate after performing a reaction. From her calculations, she knew she should have ob
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it would be 1

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The Spirit of America Goodyear blimp has a volume of 2.027×105 ft3.When the blimp is in operation, it is filled with helium gas.
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Answer: The empirical formula is C_3H_3O.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2 = 12.24 g

Mass of H_2O = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

5 0
3 years ago
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