a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>
Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :
volume NO at 1273 K and 1 atm
b. 15 L NH3 at STP ( 1mol = 22.4 L)
mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :
mass H2O(MW = 18 g/mol) :
c. mol NO at 1273 K and 1 atm :
mol ratio of NO : O2 = 4 : 5, so mol O2 :
Volume O2 at STP :
Answer: 0.0069L
Explanation:
2H2O(l) ---->O2(g) + 4H+(aq) + 4e-
no of moles= it/eF
NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)
Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)
= 0.0002798 moles= 2.798x 10 ^-4moles
Using ideal gas equation,
P V = n R T
Where, P is the pressure,
V is the volume,
n is the number of moles,
R is the gas constant, and T is the temperature
We have, 1 bar = 0.986923 atm
Substituting the values,
V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L
Volume of O2 produced = 0.0069L
Answer:
Explanation:
If l = 3, the electrons are in an f subshell.
The number of orbitals with a quantum number l is 2l + 1, so there
are 2×3 + 1 = 7 f orbitals.
Each orbital can hold two electrons, so the f subshell can hold 14 electrons.
Answer:
A.) False B.)False C.) True D.) False E.) False
Explanation:
Answer: Option (a) is the correct answer.
Explanation:
Since it is given that the mass of a jelly bean is less than the mass of a gum drop.
So, when we counted out 10 of each kind of candy and measured the mass of each kind of candy, the mass of the jellybeans would be less than the mass of the gumdrops.
This is because mass of jelly bean is less and even if we take take or more jelly beans then also their total mass will remain less than the total mass of same number of gum drops.
