NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
Boiling Point, Melting Point, Viscosity, Surface Tension. Decrease: Vapor Pressure.
Answer:
o
Explanation:
it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.
Answer:
1) 0 N
2) 8 N
Explanation:
The net force is the sum of all of the forces acting on the object.
For question 1, we can see that there is a force of 5 N acting to the right and 5 N acting to the left. If we define the right to be positive and the left to be negative, then the net force equals:
Fnet = 5N - 5N = 0 N
Therefore, the net force in question 1 is 0 N.
For question 2, the process is very similar. We want to find the sum of the forces acting on the object. In this case, there are forces of 3 N and 5 N acting to the right.
Fnet = 3 N + 5 N = 8 N
Therefore, the net force in question 2 is 8 N.
Hope this helps!
