Five mineral samples of equal mass of calcite, caco3 (mm 100.085) , had a total mass of 10.9 ± 0.1 g. what is the average mass of calcium in each sample? (assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.)

The molecular mass of calcite is 100.085 (given).

Atomic mass of calcium = 40.078 amu

Percent amount of calcium in calcite = %

The total mass of five mineral samples =  (given)

So, the mass of one sample =

and

The average mass of calcium in sample =

and

Hence, the average mass of calcium in each sample is .

a basic solution is an aqueous solution containing more oh-ions than h+ions. in other words, it is an aqueous solution with a ph greater than 7. basic solutions contain ions, conduct electricity, turn red litmus paper blue, and feel slippery to the touch.

explanation:

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !