Answer:
Explanation:
percentage abundance of third isotope = 100 - ( 78.900 + 10.009)
= 11.091 %
Atomic mass
24.1687 x .789 + 25.4830 x .10009 + 24.305 x .11091
19.069 + 2.5506 + 2.69566
= 24.3153 amu
Answer:
Tincture of iodine, iodine tincture, or weak iodine solution is an antiseptic. It is usually 2–7% elemental iodine, along with potassium iodide or sodium iodide, dissolved in a mixture of ethanol and water. Tincture solutions are characterized by the presence of alcohol.
Explanation:
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
