Using the relationship M1V1 = M2V2 where M1 and M2 are the molar concentrations (mol/L or mmol/ml) and V1 and V2 are the volumes of the solutions, we can arrive at the following answer for the given problem:
<span>15.0M (L of stock solution) = 2.35M (0.25L) *all volumes were converted to liters.
L of stock solution = (2.35*0.25)/15.0
Therefore, 0.0392L or 39.17 ml of stock solution is needed. </span>
The answer here is neutrons
The combustion of an organic compound is mostly written as,
CaHbOc + O2 --> CO2 + H2O
where a, b, and c are supposed to be the subscripts of the elements C, H, and O in the compound. Determining the number of moles of C and H in the product which is the same as that in the compound,
(Carbon, C) : (561 mg) x (12/44) = 153 mg x (1 mmole/12 mg) = 12.75
(Hydrogen, H) : (306 mg) x (2/18) = 34 mg x (1 mmole/1 mg) = 34
Calculating for amount of O in the sample,
(oxygen, O) = 255 - 153 mg - 34 mg = 68 mg x (1mmole/16 mg) = 4.25
The empirical formula is therefore,
C(51/4)H34O17/4
C3H8O1
The molar mass of the empirical formula is 60. Therefore, the molecular formula of the compound is,
C9H24O3
Answer:
Having a specific set of possible values
