A mixture contains N a H C O 3 together with unreactive components. A 1.56 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

Explanation:

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In this case, since a symbolic representation of the undergoing chemical reaction is:

In such a way, since there is 1:1 mole ratio between the sodium bicarbonate (84.01 g/mol) and carbon dioxide (44.01 g/mol) we compute the mass of reactant that was used during the reaction via stoichiometry:

Thus, the by-mass percent of pure sodium bicarbonate in the 1.56-g sample is:

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for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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