I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.
The density of water is 1kg/L. Since the density of the block is less, it will float.
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
Cause it felt like being cold
Answer:
Explanation:
moles of acetic acid = 500 x 10⁻³ x .1 M
= 5 X 10⁻³ M
.005 M
Moles of NaOH = .1 M
Moles of sodium acetate formed = .005 M
Moles of NaOH left = .095 M
pOH = 4.8 + log .005 / .095
= 4.8 -1.27875
= 3.52125
pH = 14 - 3.52125
= 10.48
Look it up, it’s not that hard.
