How many cookies are there in one mole of cookies?

A compound possessing a carboxylic acid produces a yellow solution when dissolved in diethyl ether. This yellow solution is tran

Serga [27]

Answer:

A. The top layer will be diethyl ether, and the top layer will be yellow.

Explanation:

The purpose of the addition of the saturated aqueous solution of polar solvents like sodium chloride in the liquid-liquid extraction techniques is to remove as well as separate any kind of water which may be dissolved in the ether. Water and sodium chloride are both polar and thus, they forms the bottom layer and only ether forms the top layer. The compound being organic and is colored is in the top layer with the ether.

Hence, answer - A. The top layer will be diethyl ether, and the top layer will be yellow.

6 0

3 years ago

Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat

Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

Density of aluminum (ρ): 2.7 g/cm³

Mass of aluminum (m): 81 g

Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

4 0

3 years ago

The formation of a white cloud

saw5 [17]

Most clouds are white. That's because water and ice particles that make up a cloud have just the right amount and sizes to scatter light in all possible wavelengths. When light of practically all wavelengths combine, the result is white light.

4 0

3 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

4 years ago

In a reaction vessel, 17.6 g of solid chromium(III) oxide, Cr2O3, was allowed to react with excess carbon tetrachloride in the f

notsponge [240]

Answer:

72.53% is the yield of CrCl3

Explanation:

Given

Reaction:

Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)

CCl4 is in excess and 17.6g Cr2O3 present

The reaction yields 26.6g of CrCl3

To Find:

% yields of the reaction

Also given

Molar mass of CrCl3 = 158.35g/mol

Molar mass of Cr2O3 = 152.00 g/mol

By the stoichiometry of the reaction

1 mole of Cr2O3 gives 2 moles of CrCl3

0r

1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3

= 1 52 g of Cr2O3 gives 316.70 g of CrCl3

17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3

= 36.67 g CrCl3

but actual yield is only 26.6g

so % yield is (26.6 ÷÷ 36.67) × 100

= 72.53% is the yield of CrCl3

8 0

4 years ago