Question

Чему равна масса: а) 2,24 л водорода; 6) 22,4 мл воды; в) 22,4 л хлора? Рассчитайте массу образца ВaSО4, в котором содержится 12,04-10^23 атомов кислорода.

Answer 1

Ответ:фото

Объяснение:

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Answer 2

Mass of Hydrogen:

The mass of 2.24 liters of hydrogen can be calculated using the molar mass of hydrogen, which is approximately 2 grams per mole.

To convert liters to moles, we can use 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. Assuming standard temperature and pressure (STP), which is 1 atmosphere of pressure and 273.15 Kelvin (0 degrees Celsius), we can calculate the number of moles of hydrogen.

Using the equation n = PV/RT, we can calculate the number of moles of hydrogen gas in 2.24 liters at STP. Plugging in the values, we get:

n = (1 atm) * (2.24 L) / (0.0821 L * atm / mol * K) * (273.15 K) n ≈ 0.099 moles

To calculate the mass, we multiply the number of moles by the molar mass of hydrogen:

mass = 0.099 moles * 2 g/mol mass ≈ 0.198 grams

Therefore, the mass of 2.24 liters of hydrogen is approximately 0.198 grams. [[1]]

Mass of Water:

The mass of 22.4 milliliters of water can be calculated using the density of water, which is approximately 1 gram per milliliter.

To convert milliliters to grams, we can multiply the volume by the density:

mass = 22.4 mL * 1 g/mL mass = 22.4 grams

Therefore, the mass of 22.4 milliliters of water is 22.4 grams. [[2]]

Mass of Chlorine:

The mass of 22.4 liters of chlorine gas can be calculated using the molar mass of chlorine, which is approximately 35.5 grams per mole.

To convert liters to moles, we can use 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. Assuming standard temperature and pressure (STP), which is 1 atmosphere of pressure and 273.15 Kelvin (0 degrees Celsius), we can calculate the number of moles of chlorine.

Using the equation n = PV/RT, we can calculate the number of moles of chlorine gas in 22.4 liters at STP. Plugging in the values, we get:

n = (1 atm) * (22.4 L) / (0.0821 L * atm / mol * K) * (273.15 K) n ≈ 1 mole

To calculate the mass, we multiply the number of moles by the molar mass of chlorine:

mass = 1 mole * 35.5 g/mol mass = 35.5 grams

Therefore, the mass of 22.4 liters of chlorine gas is 35.5 grams. [[3]]

Mass of BaSO4 with 12.04 x 10^23 Oxygen Atoms:

To calculate the mass of BaSO4 containing 12.04 x 10^23 oxygen atoms, we need to determine the number of moles of oxygen atoms and then use the molar mass of BaSO4 to calculate the mass.

BaSO4 is composed of one barium (Ba) atom, one sulfur (S) atom, and four oxygen (O) atoms. The molar mass of BaSO4 is the sum of the molar masses of its constituent elements.

The molar mass of oxygen is approximately 16 grams per mole. Therefore, the number of moles of oxygen atoms can be calculated as follows:

moles of oxygen = (12.04 x 10^23 atoms) / (6.022 x 10^23 atoms/mol) moles of oxygen ≈ 2 moles

Since there are four oxygen atoms in one molecule of BaSO4, the number of moles of BaSO4 can be calculated as follows:

moles of BaSO4 = (2 moles of oxygen) / (4 moles of oxygen per mole of BaSO4) moles of BaSO4 ≈ 0.5 moles

Finally, we can calculate the mass of BaSO4 by multiplying the number of moles by the molar mass of BaSO4:

mass of BaSO4 = 0.5 moles * (137.33 g/mol + 32.07 g/mol + 4 * 16 g/mol) mass of BaSO4 ≈ 0.5 moles * 233.33 g/mol mass of BaSO4 ≈ 116.67 grams

Therefore, the mass of the BaSO4 sample containing 12.04 x 10^23 oxygen atoms is approximately 116.67 grams. [[4]]