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2 years ago

2-A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11%

Chemistry

1 answer:

aleksandrvk [35]2 years ago

3 0

The molecular formula :

C₆H₁₄O₃PF

<h3>Further explanation</h3>

Given

39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine.

Required

The molecular formula

Solution

mol ratio :

C = 39.1 : 12 = 3.258

H = 7.67 : 1 = 7.67

O = 26.11 : 16 = 1.632

P = 16.82 : 31 = 0.543

F = 10.3 : 19 = 0.542

Divide by 0.542

C = 6

H : 14

O = 3

P = 1

F = 1

The empirical formula :

C₆H₁₄O₃PF

(The empirical formula)n = the molecular formula

(C₆H₁₄O₃PF)=184.1

(6.12+14.1+3.16+31+19)n=184.1

(184)n=184.1

n = 1

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3 years ago

Read 2 more answers

When adding or subtracting deelmals, how many digits should the answer contain?

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Answer:

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Explanation:

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3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

2 years ago

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa. If Po2 =

HACTEHA [7]

Answer:

A gas mixture containing oxygen, nitrogen, and carbon dioxide has a total pressure of 32.5 kPa.

<u>The pressure for oxygen is 3 kPa</u>

Explanation:

According to Dalton's Law of Partial Pressure total exerted by the mixture of non-reacting gases is equal to sum of the partial pressure of each gas.

$P_{total}=P_{1}+P_{2}+P_{3}$