Answer:
10%
Explanation:
Given data:
Actual yield = 0.13 moles
Percent yield = ?
Solution:
Balanced chemical equation.
3H₂ + N₂ → 2NH₃
First of all we will calculate the mass of given moles of ammonia.
Number of moles = mass / molar mass
Mass = 0.13 × 17 g/mol
Mass = 2.21 g
2.21 g is experimental yield of ammonia.
Hydrogen is limiting reactant because only two moles of hydrogen present.
we will compare the moles of hydrogen and ammonia from balance chemical equation.
H₂ : NH₃
3 : 2
2 : 2/3×2 = 1.33 moles
Mass of ammonia
Mass = number of moles × molar mass
Mass = 1.33 mol × 17 g/mol
Mass = 22.61 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.21 g / 22.61 g
Percent yield =9.8% which is almost 10%
The answer is transition metals because they have no specific charge except for 1 or 2 of them
Answer:
Coefficient = 1.58
Exponent = - 5
Explanation:
pH = 2.95
Molar concentration = 0.0796M
Ka = [H+]^2 / [HA]
Ka = [H+]^2 / 0.0796
Therefore ;
[H+] = 10^-2.95
[H+] = 0.0011220 = 1.122 × 10^-3
Ka = [H+] / molar concentration
Ka = [1.122 × 10^-3]^2 / 0.0796
Ka = (1.258884 × 10^-6) / 0.0796
Ka = 15.815 × 10^-6
Ka = 1.58 × 10^-5
Coefficient = 1.58
Exponent = - 5
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
