Aluminum sulfide : Al₂S₃
ratio cation : anion = 2 : 3
<h3>Further explanation</h3>
Given
Compound of Aluminium
Required
cations anions ratio
Solution
Salt can be formed from cations and anions which have their respective charges.
In the chemical compound formula these charges are crossed with each other
For aluminum it has a +3 charge
1. Aluminum carbide : Al₄C₃
ratio cation : anion = 4 : 3
2. Aluminum chloride : AlCl₃
ratio cation : anion = 1 : 3
3. Aluminum sulfide : Al₂S₃
ratio cation : anion = 2 : 3
4. Aluminum nitride : AlN
ratio cation : anion = 1 : 1
Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m
The answer:
for the monoatomic <span>selenium ions
</span> -the ion charge of selenium is 2-, so the answer is [Se]2+
as for the monoatomic phosphorus ions
-the ion charge of phosphorus is 3-, so the answer is [P]3-
Answer:
Explanation:
From the net ionic equation
Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4
Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+
Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-
Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)
