Question

КТО ХОЧЕТ 99 БАЛОВ? ТОГДА ВАМ СЮДА 1. Розрахуйте масу (г) 1,505  1023 молекул хлору: А. 0,071 Б. 1,775 В. 17,750 Г. 8,875 2Посудину, з якої заздалегідь викачали повітря, зважили. Потім заповнили газом А. Знову зважили. Газ А важить 0,02 г. Газ А відкачали і посудину заповнили СО2. Цей газ важить 0,44 г. Визначити газ А, його молекулярну массу, об’єм посудини та масу повітря. Усі виміри зроблено за нормальних умов.

Answer 1

1) Знаходимо кількість речовини хлору:
∨(Сl₂) = 1,505*10²³/6,02*10²³ = 0,25 моль
Знаходимо масу хлору:
m(Cl₂)=∨(Cl₂)*M(Cl₂) = 0,25 моль*71 г/моль = 17,75 г
Відповідь: В. 17,750
2)
Знайдемо кількість речовини вуглекислого газу СО₂
∨(СО₂) = m(CO₂)/M(CO₂) = 0,44 г/44 г/моль = 0,01 моль
За нормальних умов любий газ однаковою кількістю займає однаковий об'єм.
Розраховуємо молярну масу газу А:
М(А)=0,02 г/0,01 моль = 2 г/моль - це водень
Розраховуємо об'єм посудини:
V(посудини) = 0,01 моль*22,4 л/моль = 0,224 л
М(повітря) = 0,01 моль* 29 г/моль = 0,29 г

Answer 2

Question: Who wants 99 points? Then come here. 1. Calculate the mass (g) of 1.505 × 10^23 molecules of chlorine: A. 0.071 B. 1.775 C. 17.750 D. 8.875 2. A vessel, from which air was previously pumped out, was weighed. Then it was filled with gas A. It was weighed again. Gas A weighs 0.02 g. Gas A was pumped out and the vessel was filled with CO2. This gas weighs 0.44 g. Determine gas A, its molecular mass, the volume of the vessel, and the mass of air. All measurements were made under normal conditions.

To answer your question, let's break it down into two parts:

1. Calculate the mass of 1.505 × 10^23 molecules of chlorine: To calculate the mass of the given number of molecules of chlorine, we need to know the molar mass of chlorine (Cl2). The molar mass of chlorine is approximately 70.906 g/mol [[1]]. Using this information, we can calculate the mass as follows:

Mass of 1.505 × 10^23 molecules of chlorine = (1.505 × 10^23 molecules) × (70.906 g/mol) = 1.067 × 10^25 g

Therefore, the correct answer is B. 1.775 g [[1]].

2. Determine the properties of gas A: We are given that the initial weight of the vessel is unknown, but it is filled with gas A, which weighs 0.02 g. After pumping out gas A and filling the vessel with CO2, the weight of the vessel and CO2 is 0.44 g.

To determine the properties of gas A, we need to find its molecular mass, the volume of the vessel, and the mass of air.

- Molecular mass of gas A: The molecular mass of gas A can be calculated using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since the measurements were made under normal conditions, we can assume the pressure and temperature are constant.

We are given the mass of gas A (0.02 g) and the molar mass of gas A is unknown. Let's assume the molar mass of gas A is M g/mol. The number of moles (n) can be calculated using the formula:

n = (mass of gas A) / (molar mass of gas A) 0.02 g / M g/mol = n

Since the volume (V) and the number of moles (n) are constant, we can rewrite the ideal gas law equation as:

P1V1 = P2V2

Where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Since the pressure and temperature are constant, we can assume P1 = P2 and rewrite the equation as:

V1 = V2

Therefore, the volume of the vessel remains the same.

- Volume of the vessel: The volume of the vessel remains the same after pumping out gas A and filling it with CO2. Therefore, the volume of the vessel remains unchanged.

- Mass of air: To determine the mass of air, we need to subtract the mass of gas A from the initial weight of the vessel. However, since the initial weight of the vessel is unknown, we cannot determine the mass of air.

In summary: - The mass of 1.505 × 10^23 molecules of chlorine is 1.067 × 10^25 g, and the correct answer is B. 1.775 g. - The molecular mass of gas A, the volume of the vessel, and the mass of air cannot be determined based on the given information.

Please let me know if there is anything else I can help you with!