Bohr suggested, that there are definitive shells of particular energy and angular momentum in which an electron can revolve. It was not in Rutherford's model
They are compounds Good Luck!
They must make sure that all things pointed out are fact, not opinion.
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
Answer: 104 g
Explanation: reaction Cr2O3 + 3 H2 ⇒ 2 Cr + 3 H2O
M(Cr2O3) = 150 g/mol, so n = m/M = 1.0 mol
Number of moles of H2 should be 3.0 moles and
It is much greater (150 g / 2.016 g/mol)
1 mol Cr2O3 produces 2 mol Cr.
Mass m= 2.0 mol· 52g/mol= 104 g
