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3 years ago

Martina has a sample of an unknown substance

Chemistry

1 answer:

LiRa [457]3 years ago

7 0

I think you forgot to add some of the question

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Electrons and protons (attract/repel) each other

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3 years ago

A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

$n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH$

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

$m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO$

And consequently the moles:

$n_O=0.12molO$

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

$C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1$

Thus, the empirical formula, taken the nearest whole number is:

$C_5H_{10}O_2$

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

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3 years ago

What information do the chemical hazard label and msds have in common

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They tell you to have caution.

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3 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

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3 years ago