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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question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

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(jacemorris04)