Moles (mol) = mass (g) / molar mass (g/mol)
Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>21.7 g / 58.4 g/mol = 0.372 mol
Hence moles of NaCl in the mixture is 0.372 mol.
Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.
mole % = (moles of desired substance / Total moles of the mixture) x 100%
Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
Total moles of mixture = moles of NaCl + KCl + LiCl
Mass of KCl = 3.74 g
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g / 74.6 g/mol = 0.050 mol
Mass of NaCl = <span>9.76 g
</span>Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol
Total moles = 0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
= (0.372 mol / 0.652 mol) x 100%
= 57.06%
Hence, mixture has 57.06% of NaCl as the mole percentage.
Acids have ph less than 7 and alkali have ph more than 7
Answer:
Equilibrium constant Kc = Qc = quotient of reactant(s) and product(s)
Kc = [C]x[D]y..../[A]m[B]n..... = 0.328dm3/mol, where [C]x[D]y is the product and [A]m[B]n is the reactant(Both in gaseous states)
Explanation:
When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant (K) of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as Kc.
That a reaction quotient always assumes the same value at equilibrium can be expressed as:
Qc (at equilibrium) = Kc =[C]x[D]y…/[A]m[B]n…
This equation is a mathematical statement of the law of mass action: When a reaction has attained equilibrium at a given temperature, the reaction quotient for the reaction always has the same value.
Answer:
101.2%
Explanation:
Given:
Theoretical yield of the precipitate = 0.914 g
Actual yield of the precipitate = 0.925 g
Now, the percent yield is given as a ratio of actual yield by theoretical yield expressed as a percentage.
Framing in equation form, we have:
Now, plug in 0.925 g for actual yield, 0.914 g for theoretical yield and solve for % yield. This gives,
Therefore, the percent yield is 101.2%.
