Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole


Moles = 0.246

Volume = 280 mL = 0.280 L

Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g

Moles = 1.98
Volume = 2.6 L


Molarity = 0.762 M
Molarity = 0.76 M
Answer:
c) 2.5 mL
Explanation:
Solution
Doctors order = 0.125g
and
The liquid suspension concentration = 250 mg/5ml
= 0.250g/5ml
Or 0.05g/ml
Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml
I believe the answer is A.
M = 2 . 8 . 2
Valence Electron of M = 2
M ==> M⁺² + 2 e⁻
a. M⁺² + OH⁻ ==> M(OH)₂
b. M⁺² + PO₄⁻³ ==> M₃(PO₄)₂
The change is called melting