The method is called the displacement method.

You place some water in the graduated cylinder and measure its volume.
Then you add your object and measure the new volume.
The difference between the two volumes is the volume of your object.

6 0

3 years ago

How many CO2 molecules in 0.0189 mol of CO2?

Blizzard [7]

Answer:

option C is correct = 1.14 × 10²² molecules of CO₂

Explanation:

Given data:

Number of moles of CO₂ = 0.0189 mol

Number of molecules = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

For given question:

1 mole of CO₂ = 6.022 × 10²³ molecules of CO₂

0.0189 mol of CO₂ × 6.022 × 10²³ molecules of CO₂ / 1mol

1.14 × 10²² molecules of CO₂

Thus, option C is correct.

4 0

3 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

3 years ago