All categories

3 years ago

How many molecules of CO2 are in 132.03 grams of CO2 ?

Chemistry

1 answer:

Damm [24]3 years ago

3 0

Answer: 3.00 molecules

Explanation:

Given that:

Mass of CO2 = 132.03 grams

Number of molecules of CO2 = ?

Molar mass of CO2 = ?

For the molar mass of CO2: Atomic mass of Carbon = 12; Oxygen = 16

= 12 + (16 x 2)

= 12 + 32 = 44g/mol

Apply the formula:

number of molecules = (mass in grams/molar mass)

N = (132.03 grams/44g/mol)

N = 3.00 molecules

Thus, there are 3.00 molecules of CO2 in 132.03 grams of CO2.

You might be interested in

A lightbulb filled with argon gas has a volume of 75 mL at STP. How many moles of argon gas does the lightbulb contain?

Elan Coil [88]

Answer:

0.00335 moles

Explanation:

From the question, Using

PV = nRT................... Equation 1

Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³

Constant: R = 0.082 atm·dm³/K·mol

Substitute into equation 2

n = (1×0.075)/(273×0.082)

n = 0.075/22.386

n = 0.00335 moles

4 0

3 years ago

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

3 0

2 years ago

Which will have a greater volume: 10 grams of liquid water or 10 grams of solid water?

san4es73 [151]

The volume of 10 grams of frozen water is more than the volume of 10 grams of liquid water

5 0

3 years ago

A major textile dye manufacturer developed new yellow dye. The dye has a percent composition of 75.9 5% C, 17.72% N

AnnZ [28]

Answer:

Such molecule must have molecular formula of C15N3H15

Explanation:

Mass of carbon in such molecule

$0.7595*240_{g/mol} =182.28_{g C/mol}$

The atomic mass of carbon is 12.01 g/mol, so in 182.28 g of carbon there is 15.18 mols of carbon.

Mass of Nitrogen in such molecule

$0.1772*240_{g/mol} =37.73_{g C/mol}$

The atomic mass of nitrogen is 14.01 g/mol, so in 42.53g of nitrogen there is 3.04 mols of nitrogen.

Mass of Hydrogen in such molecule

$0.0633*240_{g/mol} =15.19 {g C/mol}$

The atomic mass of Hydrogen is 1.00 g/mol, so in 15.19 g of Hydrogen there is 15.19 mols of Hydrogen.

Such molecule must have molecular formula of C15N3H15

5 0

3 years ago

1.Oxygen (O) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and the other elements

Lady_Fox [76]

1. A. All the elements in the column have similar chemical properties.

2. Substances on the periodic table cannot be broken down into other substances and are therefore elements.

6 0

3 years ago

Read 2 more answers