All categories

3 years ago

A certain liquid has a density of 2.67 g/cm3. 30.5 ml of this liquid would have a mass of __________ kg.

1 answer:

8 0

Hey there!

Density = 2.67 g/cm³

volume = 30.5 mL

Mass = density *volume

Mass = 2.67 * 30.5

Mass = 81.435 g

Therefore:

Mass in kg :

1 kg -------------- 1000 g
?? ----------------- 81.435 g

mass = 81.435 * 1 / 1000

mass = 0.081435 kg

You might be interested in

Match the molecular shapes to the correct Lewis structures

natali 33 [55]

Answer:

1) trigonal planar

2)tetrahedral

3)trigonal pyramid

4) linear

8 0

2 years ago

Read 2 more answers

An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical

sladkih [1.3K]

Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help

7 0

3 years ago

Read 2 more answers

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

3 years ago

Help pleaseeee. over dilutions in chemistry

atroni [7]

Answer:

5 L

Explanation:

We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:

Mole of CaCl₂ = 0.5 mole

Volume = 2 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.5 / 2

Molarity = 0.25 M

Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:

Molarity of stock solution (M₁) = 0.25 M

Volume of stock solution (V₁) = 2 L

Molarity of diluted solution (M₂) = 0.1 M

Volume of diluted solution (V₂) =?

M₁V₁ = M₂V₂

0.25 × 2 = 0.1 × V₂

0.5 = 0.1 × V₂

Divide both side by 0.1

V₂ = 0.5 / 0.1

V₂ = 5 L

Thus the volume of the diluted solution is 5 L

5 0

2 years ago

What happens when the calcium burns in the oxygen. write the chemical reaction

Valentin [98]

Answer:

Calcium can be ignited and will when burning react with both oxygen and nitrogen forming calcium oxide, CaO, and calcium nitride, Ca3N2.

3 0

2 years ago