Answer:
92.72 kJ
Explanation:
2 N₂ (g) + O₂ (g) —-> 2 N₂O
According to question , one mole of N₂O requires 163.2 kJ of heat
Molecular weight of N₂O = 44 gm
25 g N₂O = 25 / 44 mole
25 / 44 mole will require 163.2 x 25 / 44 kJ
= 92.72 kJ
Answer:
The answer to your question is P2 = 0.78 atm
Explanation:
Data
Temperature 1 = T1 = 263°K Temperature 2 = T2 = 298°K
Volume 1 = V1 = 24 L Volume 2 = V2 = 35 L
Pressure 1 = P1 = 1 Pressure 2 = P2 = ?
Process
1.- To solve this problem use the Combined gas law
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (1)(24)(298) / (263)(35)
-Simplification
P2 = 7152 / 9205
-Result
P2 = 0.777
or P2 = 0.78 atm
Homeostasis is the ability to maintain stable internal conditions B is the answer
Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
