Answer:
8.1×10^-8 mols-1
Explanation:
Now we have the mass of copper sulphate produced after three days. Recall that the rate of reaction is given as;
Rate= change in the concentration of product/time
At the beginning of the reaction, there was 0 moles of copper sulphate
After 72 hours or 259200 seconds, there was 3.4g/160gmol-1 = 0.021 moles of copper sulphate.
Note that 160gmol-1 is the molar mass of copper sulphate.
Hence;
Rate of reaction= 0.021 moles /259200 seconds
Hence, the rate of reaction is 8.1×10^-8 mols-1
Rate of reaction= 8.1×10^-8 mols-1
The corect answers will be:
1) A
2) B
3) D
Hope this helped :)
The reaction between boron sulfide and carbon is given as:
2B2S3 + 3C → 4B + 3CS2
As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.
Given data:
Mass of C = 2.1 * 10^ 4 g
Mass of B = 3.11*10^4 g
Mass of CS2 = 1.47*10^5
Mass of B2S3 = ?
Now based on the law of conservation of mass:
Mass of B2S3 + mass C = mass of B + mass of CS2
Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5
Mass of B2S3 = 15.7 * 10^4 g
(2) argon. This is because Ca originally has 20 total electrons with a configuration of 2,8,8,2. When it looses its valence electrons it remains with 18 electrons total (2,8,8 config). Argon has 18 electrons total too (2,8,8 config).
