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First thing we have to do is to eliminate one of the variables, either x or y (it doesn’t matter which one). In order to achieve that, we aim to have the same co-efficient (number in front of x or y) of a variable in both equations.
In this example, we can observe that no co-efficient is the same so we have to do some manipulation first before eliminating a variable!
Multiplying 2x+5y=17 by 2, give us 4x+10y=34 (Remember to multiply by 2 all the numbers)
Now that we have the same x co-efficient in both equations we can eliminate the x variable by subtracting one equation from the other:
4x + 10y = 34
– 4x + 7y = 25
Hence, we can substitute y in one of the initial equations e.g. 2x+5y=17
2x+5(3) = 17
Therefore, the variables are x=1 and y=3see more
This is a neutralisation reaction between an acid (sulfuric acid) and a base (copper carbonate). When a carbonate reacts with an acid, the products are a salt, carbon dioxide and water.
carbonate + acid --> salt + carbon dioxide + water
ATTENTION: If you are NOT familiarised with these reactions, I advise first to revise and then try to solve the question.
First thing to do when we tackle problem like this one is to write down the balanced equation (if it is not given)
CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O
RFM (CuCO3) = 124 g mol-1
Number of moles of CuCO3 = 6.2 g / 124 g mol-1 = 0.05 mol
For every 1 mole of CuCO3, 1 mole of CO2 is produced (Using the stoichiometry of the equation)
Number of moles of CO2 = 0.05 mol
Volume of CO2 gas released = 0.05 mol * 24 dm3 = 1.2 dm3 = 12000 cm3see more