Moles of ammonium sulfate = 26.42/molar mass of (NH4)2SO4
= 26.42/132.14 = 0.19 mole.
Molarity = moles of ammonium sulfate/volume of solution
= 0.19/50x10^-3
= 3.8M
A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.
Answer:
3M
Explanation:
moles ÷ liters = molarity
4.8 ÷ 1.6 = 3M
The concentration of the original calcium ions is 0.005 M
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.
As such we have the equation;
Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)
Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol
= 0.0022 moles
Now;
1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.
Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L
= 0.005 M
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