-1 is the charge that results.
95.21 grams.
Add the atomic mass of each element present. (If there is more than one atom of an element, multiply the element's atomic mass by the number of that element's atoms.)
The molarity of a Ba(OH)2 solution required to prepare a1.0 OH- solution is calculated as follows
write the equation for dissociation of Ba(Oh)2
that is,
Ba(Oh)2 -----> Ba^2+ + 2Oh-
by use of reacting ratio between Ba(Oh)2 to Oh which is 1:2 the molarity of Ba(oh)2 = 1.0/2 = 0.5 M
Answer is: concentration is 1.3 ppm.
Parts-per-million (10⁻⁶) is present at one-millionth of a gram per gram of sample solution, for example mg/kg.
d(Cu) = 0.0013 g/L; mass concentration of copper.
d(H₂O) = 1.00 g/mL; density of water.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 1000 mL · 1 g/mL.
m(H₂O) = 1000 g ÷ 1000 g/kg.
m(H₂O) = 1 kg; mass of water.
m(Cu) = 0.0013 g · 1000 mg/g.
m(Cu) = 1.3 mg; mass of coppper.
concentration = 1.3 mg ÷ 1 kg.
concentration = 1.3 mg/kg.
concentration = 1.3 ppm.
