In the given situation, the reaction is-
NO + H2 ↔ Products
The rate of the reaction can be expressed (in terms of the decrease in the concentration of the reactants) as-
Rate = -dΔ[NO]/dt = -dΔ[H2]/dt
Now, if the concentration of NO is decreased there will be fewer molecules of the reactant NO which would decrease the its collision with H2. As a result the rate of the forward reaction would also decrease.
Ans) A decrease in the concentration of nitrogen monoxide decreases the collisions between NO and H2 molecules. the rate of the forward reaction then decreases.
Answer:
3 moles
Explanation:
To solve this problem we will use the Avogadro numbers.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms, ions or molecules in one mole of substance. According to this,
1.008 g of hydrogen = 1 mole = 6.022×10²³ atoms.
18 g water = 1 mole = 6.022×10²³ molecules
we are given 36 g of C-12. So,
12 g of C-12 = 1 mole
24 g of C-12 = 2 mole
36 g of C-12 = 3 mole
So 3 moles of C-12 equals to the number of particles in 36 g of C-12.
Answer:
I am not really sure, but I think Fr.
Explanation:
The volume of the gas that occupy at STP is 165. 28 cm^3
calculation
by use of combined gas law that is P1V1/T1=P2V2/T2, where
P1=84.6 kpa
T1=23.5 +273=296.5 K
V1=215 cm^3
At STP T= 273 K and P= 101.325 Kpa
therefore p2 = 101.325 Kpa and T2 = 272 K V2=?
by making V2 the subject of the formula V2 =T2P1V1/P2T1
V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3
