3.6*10^20 atoms* (1 mol/ (6.02*10^23 atoms))= 6.0*10^(-4) mol.
6.0*10^(-4) mol* (28.1 g/ 1 mol)= 1.7*10^(-2) g.
Note that the units cancel out so you get the answer.
The final answer is 1.7*10^(-2) g Si.
Answer:
The balanced equations for those dissociations are:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻ (aq)
H₂SO₄ (aq) → 2H⁺(aq) + SO₄⁻²(aq)
Explanation:
As a strong base, the barium hidroxide gives OH⁻ to the solution
As a strong acid, the sulfuric acid gives H⁺ to the solution
Ba(OH)₂, is a strong base so the dissociation is complete.
H₂SO₄ is considerd a strong acid, but only the first deprotonation is strong.
The second proton that is released, has a weak dissociation.
H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺ (aq) + SO₄⁻² (aq) Ka
Answer:umm I think it’s B
Explanation:
The formula for alkene is:
C nH 2n
(N is the number of carbon atoms in a molecule)
Answer:
The percent yield of NaCl is 78.7 %
Explanation:
CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2NaCl
If the NaNO₃ is determined to be in excess, the limiting reagent is the chloride. We convert the mass to moles:
31 g . 1mol / 134.45g = 0.230 moles
Ratio is 1:2, so we can make a rule of three to determine the theoretical yield
1 mol of copper (II) chloride reacts to produce 2 moles of sodium chloride
Then, 0.230 moles of CuCl₂ will react to produce (0.230 .2) /1 ) = 0.461 moles of NaCl → we convert the moles to mass → 0.461 mol . 58.45 g / 1mol = 26.9 g
To find percent yield we do → (Yield produced / Theoretical yield) . 100
(21.2 g / 26.9 g) . 100 = 78.7 %
