Answer:
Kp = 1.41 x 10⁻⁶
Explanation:
We have the chemical equation:
2 A(g) + 3 B(g)⇌ C(g)
In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):
dn= (sum moles products - sum moles reactants)
= (moles C - (moles A + moles B))
= (1 - (2+3))
= 1 - 5
= -4
We have also the following data:
Kc = 63.2
T= 81∘C + 273 = 354 K
R = 0.082 L.atm/K.mol (it is a constant)
Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:
= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶
Use Boyle's Law of Pressure: P1 x V1 = P2 x V2. Givens: P1=0.9 atm V1= 4 P2= 0.9 atm Find: V2 Equation: 0.9 atm x 4 x 4 L = 0.20 atm x V2Solve: 36 atmL= 0.20 atm x V2 18 : = V2 Short answer: The new volume is 104 ml.
Answer:
2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid
Explanation:
Write down the equation in the beginning with reactants and products:
NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Now try to balance it. Try with Na first:
2NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
Check if H atoms are also balanced. They are. That means our final reaction is:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
2 Moles of NaOH reacts with 1 mole of H₂SO₄
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100
The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams
The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams
