northern america is the correct answer i’m pretty sure
Answer:
I am not really sure, but I think Fr.
Explanation:
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 : The partial pressure of the
in the tank in psia is, 32.6 psia.
Explanation :
As we are given 75 %
and 25 %
in terms of volume.
First we have to calculate the moles of
and
.


Now we have to calculate the mole fraction of
.


Now we have to calculate the partial pressure of the
gas.


conversion used : (1 Kpa = 0.145 psia)
Therefore, the partial pressure of the
in the tank in psia is, 32.6 psia.
Answer:
B. Below the equilibrium price
Explanation:Price ceiling is the legal price,impose by The Government or Regulatory agencies above which a product should not be sold. It is also a price control mechanism through which the Government and other Regulatory agencies check the excesses of producers and the middle men(Whole sellers and retailers). The price ceiling has to be fixed below the equilibrium price for it to be binding, because businesses are interested mainly for profit maximization.