PH= -log [H+]
= - (log 0.034)
= - (-1.5)
= 1.5
Concentration of Na2CO3 = 0.600 M
Volume of Na2CO3 = 30.0 ml = 0.030 L
Molarity = moles of solute/volume of solution
Now,
moles of Na2CO3 = M * V = 0.600*0.030 = 0.018 moles
Na2CO3 ↔ 2Na^+ + CO3^2-
As per stoichiometry:
1 mole of Na2CO3 produces 2 moles of Na+ ions
Therefore, 0.018 moles of Na2CO3 would dissociate to give: 2*0.018 = 0.036 moles of Na+ ions.
Now,
1 mole of Na+ has 6.023 * 10^23 ions
therefore, 0.036 moles of Na+ would correspond to:
= 0.036 * 6.023 * 10^ 23 = 2.17 *10^22 Na+ ions
Well, all of this we owe it to Bohr who analyzed the atomic emission spectrum of hydrogen and he could probe matematically that it was a result of movement of e- from an especific energy level to a lower one. The understanding of levels of energy took to the development of the atomic theory
Answer:
Theoretical yield = 3.52 g
Percent yield =65.34%
Explanation:
Given data:
Mass of HgO = 46.8 g
Theoretical yield of O₂ = ?
Percent yield of O₂ = ?
Actual yield of O₂ = 2.30 g
Solution:
Chemical equation:
2HgO → 2Hg + O₂
Number of moles of HgO = mass/ molar mass
Number of moles of HgO = 46.8 g / 216.6 g/mol
Number of moles of HgO = 0.22 mol
Now we will compare the moles of HgO with oxygen.
HgO : O₂
2 : 1
0.22 : 1/2×0.22 = 0.11 mol
Theoretical yield:
Mass of oxygen = number of moles × molar mass
Mass of oxygen = 0.11 mol × 32 g/mol
Mass of oxygen = 3.52 g
Percent yield :
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.30 g/ 3.52 g × 100
Percent yield =65.34%
