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3 years ago

If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?

1 answer:

7 0

The combustion reaction is as expressed,

CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g.

moles of C = 0.86 g C / 12 g = 0.0713
moles of H = 0.14 g H / 1 g = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.

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3 years ago

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Zielflug [23.3K]

Answer:

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Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

$\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}$

Step 2. Convert moles to grams

$\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}$

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3 years ago

What's a vacuum exactly

stealth61 [152]

Answer:

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Explanation:

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2 years ago

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Answer:

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Explanation:

Here,

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2 years ago

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