Chemistry Geniuses only please

Test Preparation AP Test Preparation
1

<p>[1] Which of the following solutions contains the largest number of ions?</p>

<p>a: 500 ml of .10M CaCl2
b: 500 ml of .10M FeCl3
c: 700 ml of .20M NaOH
d: 400 ml of .100M Al(NO3)3
e: 600 ml of .200M AlCl3</p>

<p>please show work if you can </p>

<p>[2] Write the equation and Net Ionic Equation: (no need to balance)</p>

<p>excess ammonia is added to solution of copper(II)chloride</p>

<p>[3] How many mL of .25M BaCl2 is required to precipitate all the sulfate ion from 10mL of solution containing 100g of Na2SO4 per liter?</p>

<p>thanx in advance guys....</p>

2

<p>1a: 0.5L*.1mol/L CaCl2 = 0.05mol CaCl2
CaCl2 —> Ca (2+) + 2Cl-<br>
0.05 mol CaCl2 X 3 mol ions/mol CaCl2 = 0.15 mol of ions</p>

<p>same procedure for all options you get
b)0.20 mol ions
c) 0.28 mol ions (NaOH—> Na + OH)
d)0.16 mol ions(Al (NO3)3 —> Al + 3NO3)
e)0.48 mol ions
ANSWER E</p>

<p>2) NH3 added to CuCl2(aq)
2NH3(aq) + Cu(2+)aq + –> 2NH4(+)aq + Cu(s)
spectator ion: Cl- (aq)</p>

<p>3) 100g Na2SO4 X mol/142g = 0.704 mol sodium sulfate
[Na2SO4]= 0.704mol/L
Molecular Eq: BaCl2 (aq) + Na2SO4 (aq) –> BaSO4 (s) + 2NaCl
Ksp = 1.1X10^-10 (since Ksp is so small, assume that BaSO4 is infinitely insoluble so we can estimate that 0.704M Ba(2+) from BaCl2 is required
0.01L Na2SO4 used X 0.704M = 0.0704 mol sodium sulfate
look at stoich. ratios we need same mols of barium choloride
(x litres)(0.25M) = 0.0704
x=0.2816L
=281.6 mL
= 2.8X10^2 mL (sig figs)</p>

3

<p>^2) is incorrect. When excess concentrated ammonia is added to Cu2+, a complex ion is formed.
2NH3 + Cu2+ –> Cu(NH3)4(2+)</p>

4

<p>thanx…you guys are awesome!</p>