Question

Помогите решить задачу по химии,пожалуйста очень нужно.Я вам буду очень благодарна!!!Рассчитайте массу 2,24 л оксида углерода(IV).Сколько это составляет молей и молекул

Answer 1

CH4 + 2O2 = CO2 + 2H2O
n(CH4) = V/Vm = 2,24/22,4 = 0,1 моль
n(CO2) = n(CH4) = 0,1 моль по уравнению реакции
m(CO2) = n*M = 0,1*44 = 4,4 г
n(H2O) = 2*n(CH4) = 2*0,1 = 0,2 моль по уравнению реакции
m(H2O) = n*M = 0,2*18 = 3,6 г
Ответ: 4,4 г СО2 и 3,6 г Н2О

Answer 2

N(CO2)=V(CO2)/Vм=2,24/22,4=0,1 моль
N(CO2)=n(CO2)*N(авогадро)=0,1*6,02*10^23=6,02*10^22 молекул
m(CO2)=n(CO2)*M(CO2)=0,1*44=4,4 г

Answer 3

I can help you with your chemistry problem.

The question is: Calculate the mass of 2.24 liters of carbon dioxide (IV). How many moles and molecules does this amount to?

To solve this problem, we need to use the following steps:

1. Find the molar mass of carbon dioxide (IV), which is also known as carbon tetrachloride (CCl₄). The molar mass is the sum of the atomic masses of the elements in the compound. The atomic masses of carbon and chlorine are 12 g/mol and 35.5 g/mol, respectively. Therefore, the molar mass of CCl₄ is:

$$12 + 4 \times 35.5 = 154 \text{ g/mol}$$

2. Find the number of moles of CCl₄ in 2.24 liters. To do this, we need to use the ideal gas law, which states that:

$$PV = nRT$$

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Assuming that the gas is at standard temperature and pressure (STP), which means that P = 1 atm and T = 273 K, we can simplify the equation as:

$$V = nRT/P$$

Substituting the values of R = 0.0821 L atm/mol K and V = 2.24 L, we get:

$$2.24 = n \times 0.0821 \times 273 / 1$$

Solving for n, we get:

$$n = 2.24 / (0.0821 \times 273) = 0.0998 \text{ mol}$$

3. Find the mass of CCl₄ in 0.0998 mol. To do this, we need to multiply the number of moles by the molar mass. We get:

$$0.0998 \times 154 = 15.37 \text{ g}$$

4. Find the number of molecules of CCl₄ in 0.0998 mol. To do this, we need to multiply the number of moles by the Avogadro's number, which is 6.022 x 10²³ molecules/mol. We get:

$$0.0998 \times 6.022 \times 10^{23} = 6.01 \times 10^{22} \text{ molecules}$$

Therefore, the answer is:

- The mass of 2.24 liters of carbon dioxide (IV) is 15.37 g. - The number of moles of carbon dioxide (IV) is 0.0998 mol. - The number of molecules of carbon dioxide (IV) is 6.01 x 10²² molecules.

I hope this helps you understand the problem better. If you need more help with chemistry, you can try using [Mathway](https://www.mathway.com/ru/Chemistry), a free online tool that can solve various chemistry problems with step-by-step explanations.