= 1 : 1
Therefore, the equation is Cu + O → CuO.
But Oxygen is diatomic, so we have 2Cu+O2 → 2CuO
2. Calculate the;
- mass of anhydrous sodiumtrioxocarbonate (IV) present in a 300cm3 of 0.1M;
- Number of Na2CO3 particles present in the solution (Na =23, C=12, O=16)
Solutions
a. Molar concentration of Na2CO3 =0.1M
Molar mass of Na2CO3 = 106gmol-1
Mass concentration = Molar concentration X Molar mass
=0.1 X 106
=10.6gdm-3
i.e 1000cm3 of 0.1M solution contain 10.6g of Na2CO3
Therfore,300cm3 of 0.1M solution will contain
= 300 X 10.6/1000
=3.18g of Na2CO3
b. Number of Na2CO3 particles
=Molar concentration X 6.02 X 1023
=0.1 X 6.02 1023
=6.02 X 1022
Now, 100cm3 of 0.1M solution contains 6.02X1022 Na2CO3 particles
Therefore, 300cm3 of 0.1M solution will contain 300X6.02 X1022/1000
= 1.81 X 1022 particular of Na2CO3.
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