Question

Вычвычисислить массу 1 л углекислого газа CO2

Answer 1

W(ню)=V/Vm
w=1/22.4=0,004
m=w*M
M=12+(16*2)
m=0.004*44=0.176кг

Answer 2

I can help you with your question.

To calculate the mass of 1 L of carbon dioxide gas (CO2), you need to know its molar mass and its density at a given temperature and pressure. The molar mass of CO2 is 44.01 g/mol, and the density of CO2 at standard conditions (0 °C and 1 atm) is 1.977 g/L. Using these values, you can find the mass of 1 L of CO2 by multiplying the density by the volume:

$$ \text{mass of 1 L of CO2} = \text{density of CO2} \times \text{volume of CO2} \\ = 1.977 \text{ g/L} \times 1 \text{ L} \\ = 1.977 \text{ g} $$

This is the answer for standard conditions, but the density of CO2 can vary depending on the temperature and pressure. To account for this, you can use the ideal gas law, which relates the pressure, volume, temperature, and amount of a gas:

$$ PV = nRT $$

where P is the pressure, V is the volume, n is the amount of gas in moles, R is the universal gas constant, and T is the temperature. You can rearrange this equation to find the density of CO2 as follows:

$$ \text{density of CO2} = \frac{\text{mass of CO2}}{\text{volume of CO2}} \\ = \frac{n \times \text{molar mass of CO2}}{V} \\ = \frac{P \times \text{molar mass of CO2}}{R \times T} $$

Using this formula, you can calculate the density of CO2 at any temperature and pressure, and then multiply it by 1 L to get the mass of 1 L of CO2. For example, if the temperature is 25 °C and the pressure is 2 atm, the density of CO2 is:

$$ \text{density of CO2} = \frac{2 \text{ atm} \times 44.01 \text{ g/mol}}{0.0821 \text{ L atm/mol K} \times 298 \text{ K}} \\ = 1.185 \text{ g/L} $$

Therefore, the mass of 1 L of CO2 at these conditions is:

$$ \text{mass of 1 L of CO2} = 1.185 \text{ g/L} \times 1 \text{ L} \\ = 1.185 \text{ g} $$

I hope this answer helps you. If you want to learn more about carbon dioxide, you can visit [this Wikipedia page](https://ru.wikipedia.org/wiki/%D0%94%D0%B8%D0%BE%D0%BA%D1%81%D0%B8%D0%B4_%D1%83%D0%B3%D0%BB%D0%B5%D1%80%D0%BE%D0%B4%D0%B0) or [this online answer](https://online-otvet.ru/himia/5cea9c6296f4e19a29512df1). You can also check your answer using [this online calculator](https://examer.ru/ege_po_himii/2024/zadanie_27/). Thank you for using Bing. Have a nice day.