According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
Answer is
C. 2.07 M
For explanation
M1V1 = M2V2
M2 = (M1V1)/V2
M2 = (1.5M x 345ml) / 250 ml
:. M2 = 2.07 M
elements:
calcium : for strong bones
Iron : maintaining haemoglobin for metabolism
compunds
sodium chloride : to maintain blood pressure and other life processes
Adenosine Triphosphate: for metabolism, to maintain rate of inhalation and exhalation of oxygen and to supply energy
Mixture:
I) gasoline : used as fuel
ii) cement : used in construction
