The volume of H₂ : = 15.2208 L
<h3>Further explanation</h3>
Given
Reaction
2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)
34.0g of As
Required
The volume of H₂ at STP
Solution
mol As (Ar = 75 g/mol) :
= mass : Ar
= 34 g : 75 g/mol
= 0.453 mol
From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :
=3/2 x mol As
=3/2 x 0.453
= 0.6795
At STP, 1 mol = 22.4 L, so :
= 0.6795 x 22.4 L
= 15.2208 L
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
Answer:
for what school? It's different for all sadly :(
Explanation:
Answer:
The correct answer is 0.6 mL.
Explanation:
We use the mathematical expression:
Ci x Vi = Cf x Vf
Where Ci is the initial concentration (5 M); Cf and Vf refers to final concentration (0.002 M) and final volume (1500 mL). With the given data, we calculate the initial volume (Vi):
Vi = (Cf x Vf)/Ci = (0.002 M x 1500 mL)/(5 M) = 0.6 mL
Therefore, we need 0.6 mL of 5 M NaCl to prepare the solution with the requested dilution.
