Question

Сколько атомов азота и водорода содержит 44,8 л аммиака (н.у)?

Answer 1

V(NH4)=44.8 л. V=n*Vm ,Vm=22.4л/моль, m=n*M
n=44.8л/22.4 л/моль=2моль
M(NH4)=(14+1*4)г/моль=18г/моль
m=2 моль * 18г/моль =36г.

Answer 2

Амииак формула NH3
1 моль как видим сожержиь 3 атома водорода и 1 атом азота.
3NH3
3 моль будет содержать 3 атома азота и 9 атома водорода

Answer 3

Calculation of the Number of Nitrogen and Hydrogen Atoms in 44.8 L of Ammonia at Standard Conditions

To determine the number of nitrogen and hydrogen atoms in 44.8 L of ammonia at standard conditions, we need to use the ideal gas law and the molar ratio of ammonia.

The ideal gas law equation is given by:

PV = nRT

Where: - P is the pressure of the gas (at standard conditions, it is 1 atm) - V is the volume of the gas (44.8 L in this case) - n is the number of moles of the gas - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature of the gas (at standard conditions, it is 273.15 K)

We can rearrange the equation to solve for the number of moles (n):

n = PV / RT

Now, we need to determine the molar ratio of ammonia (NH3). From the chemical formula of ammonia, we know that it contains one nitrogen atom (N) and three hydrogen atoms (H).

Therefore, the molar ratio of nitrogen to ammonia is 1:1, and the molar ratio of hydrogen to ammonia is 3:1.

To calculate the number of moles of ammonia, we can use the molar volume of a gas at standard conditions, which is 22.4 L/mol.

moles of ammonia = volume of ammonia / molar volume

moles of ammonia = 44.8 L / 22.4 L/mol

moles of ammonia = 2 mol

Since the molar ratio of nitrogen to ammonia is 1:1, the number of moles of nitrogen is also 2 mol.

Since the molar ratio of hydrogen to ammonia is 3:1, the number of moles of hydrogen is 3 times the number of moles of ammonia.

moles of hydrogen = 3 * moles of ammonia

moles of hydrogen = 3 * 2 mol

moles of hydrogen = 6 mol

Now, we can calculate the number of atoms of nitrogen and hydrogen using Avogadro's number, which is 6.022 x 10^23 atoms/mol.

number of nitrogen atoms = moles of nitrogen * Avogadro's number

number of nitrogen atoms = 2 mol * 6.022 x 10^23 atoms/mol

number of nitrogen atoms = 1.2044 x 10^24 atoms

number of hydrogen atoms = moles of hydrogen * Avogadro's number

number of hydrogen atoms = 6 mol * 6.022 x 10^23 atoms/mol

number of hydrogen atoms = 3.6132 x 10^24 atoms

Therefore, 44.8 L of ammonia at standard conditions contains approximately 1.2044 x 10^24 nitrogen atoms and 3.6132 x 10^24 hydrogen atoms.

Please note that the calculations provided are based on the assumption that the ammonia gas behaves ideally at standard conditions.