An electron in the 3s orbital. The order of electron orbital energy levels starting from lowest to highest is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
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
Using charles law
v1/t1=v2/t2
v1=1l
v2=1.1l
t2=255+273=528
t1=?
1/t1=1.1/528
cross multiply
1.1t1=528 divide both sides by 1.1
t1=528/1.1
t1=480k or 207celcius
I believe the answer would be energy
