Ask question

Ask question

All categories

2 years ago

10

Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance travelled by the solvent f

ront is 2.00 cm, and the distance the ion is from the solvent front is 0.20 cm?
Hint: Be careful with how Snape worded this question.

1 answer:

bija089 [108]2 years ago

7 0

Answer:

The retention factor of an ion is 0.10 .

Explanation:

Retention factor is defined as ratio of distance of distance traveled by solute to the distance traveled by solvent on chromatogram.

$R_f=\frac{d_{solute}}{d_{solvent}}$

We have:

$d_{solute}=d_{ion}=0.20 cm$

$d_{solven}=2.00 cm$

The retention factor of an ion :

$R_f=\frac{0.20 cm}{2.00 cm}=0.10$

You might be interested in

Which of these pumpkins has the most<br> thermal energy? Explain your answer.

VLD [36.1K]

Answer:

big

Explanation:

5 0

2 years ago

The density of gold is 19.3 g/cm3. what volume would 131.2 g of gold have?

Luda [366]

Volume would be 6.79792764 cm³. Just divide 131.2 by 19.3

ρ=Mass/Volume
Volume= mass/density

5 0

2 years ago

When 83.3 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, h

Nataly [62]

<span>2.0 moles. (83.3g / 208 [grams/moles] *5 [from balancing] = around 2 </span>

6 0

3 years ago

Read 2 more answers

Convertible bonds are examples of: a. Securities that have characteristics of both debt and equity b. Non-recurring items c. Ass

iren2701 [21]

Answer:

a. Securities that have characteristics of both debt and equity

Explanation:

A convertible bond is a debt security that can be converted into a certain number of shares and this can be done during specific times in the life of the bond. According to this, convertible bonds are examples of securities that have characteristics of both debt and equity as they are a type of debt that pays interest but can be turned into shares.

5 0

2 years ago

If 9.8g water is used in electrolysis, what is the percent yield if 5.6g of oxygen was

ANEK [815]

Answer:

Approximately $64\%$ .

Explanation:

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The actual yield of $\rm O_2$ was given. The theoretical yield needs to be calculated from the quantity of the reactant.

Balance the equation for the hydrolysis of water:

$\rm 2\, H_2O \, (l) \to 2\, H_2\, (g) + O_2\, (g)$ .

Note the ratio between the coefficient of $\rm H_2O\, (g)$ and $\rm O_2\, (g)$ :

$\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ .

This ratio will be useful for finding the theoretical yield of $\rm O_2\, (g)$ .

Look up the relative atomic mass of hydrogen and oxygen on a modern periodic table.

$\rm H$ : $1.008$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm H_2O$ and $\rm O_2$ :

$M(\mathrm{H_2O}) =2\times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{O_2}) =2\times 15.999 = 31.998\; \rm g \cdot mol^{-1}$ .

Calculate the number of moles of molecules in $9.8\; \rm g$ of $\rm H_2O$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{9.8\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \approx 0.543991\;\rm g \cdot mol^{-1}$ .

Make use of the ratio $\displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} = \frac{1}{2}$ to find the theoretical yield of $\rm O_2$ (in terms of number of moles of molecules.)

$\begin{aligned} n(\mathrm{O_2}) &= \displaystyle \frac{n(\mathrm{O_2\, (g)})}{n(\mathrm{H_2O\, (aq)})} \cdot n(\mathrm{H_2O}) \\ &\approx \frac{1}{2} \times 0.543991\; \rm mol \approx 0.271996\; \rm mol \end{aligned}$ .

Calculate the mass of that approximately $0.271996\; \rm mol$ of $\rm O_2$ (theoretical yield.)

$\begin{aligned}m(\mathrm{O_2}) &= n(\mathrm{O_2}) \cdot M(\mathrm{O_2}) \\ &\approx 0.271996\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \approx 8.70331 \; \rm g \end{aligned}$ .

That would correspond to the theoretical yield of $\rm O_2$ (in term of the mass of the product.)

Given that the actual yield is $5.6\; \rm g$ , calculate the percentage yield:

$\begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \\ &\approx \frac{5.6\; \rm g}{8.70331\; \rm g} \times 100\% \approx 64\%\end{aligned}$ .

4 0

2 years ago