<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=
106.16×1000
17.12
=0.00016moles
Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO
2
=
44.01×1000
56.77
=0.0013
Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H
2
O==
18.02×1000
14.53
=0.0008
Moles ratios
\frac{0.0013}{0.0008}=1.625
0.0008
0.0013
=1.625
\frac{0.0008}{0.0008}=1
0.0008
0.0008
=1
Hence molecular fomula
The empirical formula is C 4H 5.
The molecular formula C8H10
Answer:
metal Atom
Explanation:
every transition metal atom are responsible for the flame color. Some metal are also confirmed by flame test.
