Amixture was found to contain 1.05g of sio2, 0.69g of cellulose, 1.82g of calcium carbonate. the percentage of calcium carbonate is in the is [x]%. note: round to the nearest percent.

Mass mixture = mass SiO2 + mass cellulose + mass CaCO3 = 
mass mixture  = 1.05 g + 0.69g + 1.82g = 3.56g

mass CaCO3 = 1.82 g

%CaCO3 = mass CaCO3/mass mixture * 100 = 1.82/3.56 * 100 = 51.123%  = 51%

The porcentage of calcium carbonate in the mixture is calculated as the weight of calcium carbonate divided by the total wieght of the mixture times 100

% calcium carbonate = 1.82 g * 100 / [ 1.05 g + 0.69 g + 1.82 g] = 182 / 3.56 = 51.12 %

It dont even matter because at the end of thre day we gone life like it ius s

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.