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3 years ago

A sample of a compound contains 60.0 g C and 5.05 g H. It’s molar mass is 78.12 g/mol. What is the compound’s molecular formula?

Chemistry

1 answer:

Natalka [10]3 years ago

5 0

Answer:

C6H6

Explanation:

C : 60.0 g*1 mol/12 .01g = 5.00 mol

H: 5.05 g * 1 mol/1.01 g= 5.10 mol

C:H = 5.00 mol : 5.10 mol = 1:1

CH is an empirical formula

Molar mass (12.01 +1.01)=13.02 g/mol

78.12g/mol /13.02 g/mol =6

Molecular formula is the empirical that was taken 6 times (CH)6 = C6H6

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er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The

liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

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3 years ago

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Talja [164]

Answer:

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2 years ago

Which part of earth absorbs the most sunlight?

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2 years ago

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Which state(s) of matter do not have a definite shape nor volume? A. Liquids only B. Liquids and gases C. Gases only D. Gases an

Evgen [1.6K]

<h2>Answer:</h2>

C. Gases only

<h3>Explanation:</h3>

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3 years ago

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an

mafiozo [28]

Answer:

$\rm K$ : approximately $22.0\%$ .

$\rm Cl$ : approximately $19.9\%$ .

$\rm O$ : approximately $27.0\%$ .

$\rm Al$ : $31.0\%$ .

Explanation:

Consider a $100\; \rm g$ sample. There would be $69.0\% \times 100\; \rm g = 69.0\; \rm g$ of $\rm KClO_{3}$ and $100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g$ of $\rm Al$ .

Out of that $69.0\; \rm g$ of $\rm KClO_{3}$ :

$31.9\%$ (by mass) would be $\rm K$ : $31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g$ .

$28.9\%$ (by mass) would be $\rm Cl$ : $28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g$ .

$39.2\%$ (by mass) would be $\rm O$ : $39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g$ .

Overall, the composition (by mass) of each element in this mixture would be:

$\rm K$ :

$\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}$ .

$\rm Cl$ :

$\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}$ .

$\rm O$ :

$\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}$ .

$\rm Al$ :

$\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}$ .

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2 years ago