Answer:
We have to add 17.2 grams of aluminium bromide
Explanation:
Step 1: Data given
Molarity of the aluminum bromide solution = 0.215 M
Volume = 300 mL = 0.300 L
Molar mass aluminium bromide = 266.69 g/mol
Step 2: Calculate moles Aluminium bromide
moles AlBr3 = volume * molarity
Moles AlBr3 = 0.300 L * 0.215 M
Moles AlBR3 = 0.0645 moles
Step 3: Calculate mass aluminium bromide
Mass aluminium bromide = moles AlBr3 * molar mass AlBr3
Mass aluminium bromide = 0.0645 moles * 266.69 g/mol
Mass aluminium bromide = 17.2 grams
We have to add 17.2 grams of aluminium bromide
Answer:
The answer is B (sodium is more reactive)
Explanation:
Uranium fission is a property
In a neutral compound sum of charges should be equal 0,
+3*2-2*3=0,
so ratio of ions n(Al³)⁺:n(S²⁻)=2:3
