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Answer:
amount of silver chloride required is 0.015 moles or 2.1504 g
Explanation:
0.1M AgCL means 0.1mol/dm³ or 0.1mol/L
1L = 1000mL
if 0.1mol of AgCl is contained in 1000mL of solution
then x will be contained in 150mL of solution
cross multiply to find x
x = (0.1*150)/1000
x= 0.015 moles
moles of silver chloride present in 150 mL of solution is 0.15 moles
To convert this to grams, simply multiply this value by the molar mass of silver chloride
molar mass of silver chloride AgCl =107.86 + 35.5
=143.36 g/mol
mass of AgCl = moles *molar mass
=0.015*143.36
=2.1504g
=
Explanation:
Molarity = mol/L or g/L
Data:
Mass>80.0g
Mr of CaCl2>111.1g/mol
V>500mL
Convert mL to Litres;. mole=g/Mr
1L=1000mL. =80g/111.1g/mol
x=500mL. =0.720072007mol
x=0.5L
Molarity= mole/volume
=0.720072007mol/0.5L
=1.440144014mol/L
=1.4401mol/L
Hope this helps. Depending on the question you can also find it in g/L but mol/L is safer.
<span>(NH4)2CO3 -> 96.09 g/mol
(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate
In this conversion, the 'grams' unit is crossed out because it is in both the numerator and the denominator, which leaves the 'mol' unit left.
Looking at the formula (NH4)2CO3, you can look at it as if it were:
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3
For every 1 mol of ammonium carbonate, you have 1 mol of carbonate ions and 2 moles of ammonium ions.
(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion)+(0.363981 mol ammonium ion) </span>
