Mass = number of mol x molar mass
Mass = 0.28mol x 55.8g/mol
Mass = 15.624g
Answer:
0.007 mol
Explanation:
We can solve this problem using the ideal gas law:
PV = nRT
where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.
Keep in mind that when we collect a gas over water we have to correct for the vapor pressure of water at the temperature in the experiment.
Ptotal = PH₂O + PO₂ ⇒ PO₂ = Ptotal - PH₂O
Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.
P H₂O = 23.8 mmHg x 1 atm/760 mmHg = 0.031 atm
V = 1750 mL x 1 L/ 1000 mL = 0.175 L
T = (25 + 273) K = 298 K
PO₂ = 1 atm - 0.031 atm = 0.969 atm
n = PV/RT = 0.969 atm x 0.1750 L / (0.08205 Latm/Kmol x 298 K)
n = 0.007 mol
Answer:
For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Explanation:
The computation of the mass of each element is given below:
As we know that
A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol
Now we multiply each of above by the molar masses
For N
= 14.0 g/mol × 2
= 28.0 gN
For H
= 1.0 g/mol × 4
= 4.0 gN
ANd, for O
= 16.0 g/mol × 3
= 48.0 gN
Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively
Answer:
D
Explanation:
The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.
The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.
Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.
