Answer:
O H C
Moles in 100g 3.33 6.65 3.33
Ratio 1.00 2.00 1.00
Possible empirical formula = 
Answer:

Explanation:
When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.
So:
24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O
The next step is to divide each mass by their molar mass to convert your grams to moles.
24.42/40.08 = 0.6092 mol
17.07/14.01 = 1.218 mol
58.85/15.99 = 3.680 mol
Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.
0.6092 mol/0.6092 mol = 1
1.218 mol/0.6092 mol = 2
3.680 mol/0.6092 mol = 6
So the empirical formula is 
Answer: hello some part of your question is missing below is the missing part
when H₂O and H₂O₂ is added to Mn(OH)₂(s) and put in water bath to dissolve
answer : attached below
Explanation:
When Mn²⁺ ions are separated from the mixture, attached below are the requires reaction equations that shows the process of separation.
Mn²⁺ ions are separated to the right of the reaction equations
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.
Answer:
35
Explanation:
because it's is still the same amount even if you freeze it