On calcination 80g of a marble were obtained 42 g of calcium oxide. The degree of purity of marble?
There are 6 ways you can draw the two marbles to include one red:
RB,RY,RW,BR,YR,WR
There are 4P2 = 12 ways to draw any two marbles
So, P(one red) = 6/12 = 1/2
CaCO3>>CaO + CO2 is the reaction I assume you are thinking on.
80gmarble*Zpercent*1moleCaCO3/100gCaCO3*ImoleCaO/1moleCaCO3*56gCaO/1moleCaO=42gCaO
solve for Z.
Z=42*100/80*56=93.75 percent.
check my work.
thank you bobpursley
To find the degree of purity of the marble, we need to calculate the percentage of calcium oxide in the marble.
The formula to calculate the percentage is:
Percentage = (Mass of the component / Total mass of the substance) x 100
In this case, the mass of calcium oxide is 42 g, and the total mass of the marble is 80 g.
So, the percentage of calcium oxide in the marble is:
Percentage = (42 g / 80 g) x 100
Calculating, we have:
Percentage = 0.525 x 100
Percentage ≈ 52.5%
Therefore, the degree of purity of the marble is approximately 52.5%.