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Barondeau Chem 119 9:35 am class Quiz 9
1.
You mix 55 mL of 1.50
M
silver nitrate (AgNO
3
) with 35 mL of 1.47
M
sodium chloride (NaCl).
Determine the mass (with 2 significant figures) of precipitated silver chloride (AgCl; 143.32
g/mol) that would form (2 points).
AgNO
3
(aq) + NaCl(aq)
AgCl(s) + NaNO
3
(aq)
Ag
+
(aq) + NO
3
-
(aq) + Na
+
(aq) + Cl
-
(aq)
AgCl(s) + Na
+
(aq) + NO
3
-
(aq)
Remove spectator ions
Ag
+
(aq) + Cl
-
(aq)
AgCl(s)
Limiting reagent problem
(1.50 moles AgNO
3
/ L)(0.055 L) = 0.0825 moles AgNO
3
or Ag
+
(1.47 moles NaCl / L)(0.035 L) = 0.05145 moles NaCl or Cl
-
Cl
-
is limiting reagent
(0.05145 moles Cl
-
)(1 mol AgCl / 1 mol Cl
-
)(143.32 g AgCl / mol) = 7.3738 or 7.4 g
2.
You have 51.5 mL of a 2.80
M
solution of Na
2
CrO
4
(
aq
). You also have 112 mL of a 2.50
M
solution of AgNO
3
(
aq
). Calculate the concentration of Na
+
after the two solutions are mixed.
Report the answer with 3 significant figures (2 points)
Determine number of moles Na+ and then divide by total volume.
(2.80 moles Na
2
CrO
4
/ L)(0.0515 L)(2 Na
+
/Na
2
CrO
4
) = 0.2884 moles Na
+
Total volume = 51.5 mL + 112 mL = 163.5 mL
Concentration = (0.2884 moles / 0.1635 L) = 1.76 M Na
+
3.
The hydronium ion concentration in a solution was measured to be 4.8
10
-6
M.
a.
What is the pH of this solution?
Report the answer with 2 significant figures (1 point).
pH = -log [H
+
]
pH = -log (4.8
10
-6
M) = 4.7
b.
Is this an acidic or basic solution (1 point)?
At room temperature: pH below 7 is acidic.
pH above 7 is basic.
Acidic
7.4 g
1.76 M
5.3
4.
Write a net ionic equation for the reaction of the weak acid HF with the strong base KOH (2
points)
HF(aq) + KOH(aq)
KF(aq) + H
2
O(
l
)
HF(aq) + K
+
(aq) + OH
-
(aq)
K
+
(aq) + F
-
(aq) + H
2
O(
l
)
Remove spectator ions
HF(aq) + OH
-
(aq)
F
-
(aq) + H
2
O(
l
)
5.
Calculate the mass of NaOH (39.997 g/mol) needed to completely neutralize 58.3 mL of a 0.240
M solution of the monoprotic acid HNO
3
. Report the answer with 3 significant figures (2 points).
NaOH(aq) + HNO
3
(aq)
NaNO
3
(aq) + H
2
O(
l
)
(0.240 moles HNO
3
/ L)(0.0583 L) = 0.013992 moles HNO
3
(0.013992 moles HNO
3
)(1 mol NaOH / 1 mol HNO
3
)(39.997 g NaOH / mol) = 0.559638 g
HF(aq) + OH
-
(aq)
F
-
(aq) + H
2
O(
l
)
0.560 g
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ANSWER
RESET
*( )
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10
45
1
210.34
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前。
1 Normal
1 No Spac... Heading 1
Heading 2
Title
Paragraph
A sample of potassium bicarbonate (KHCO3) was reacted with HCl to produce KCl. The following
Styles
7
data was collected.
Mass of beaker 28.43 g
Mass of beaker + KHCO3 36.95 g
Volume of HC1 12.0 mL
Concentration of HCl 9.0 M
Mass of beaker +residue (KCI) 33.82 g
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Mass of KHCO3
Mass of residue
Mass of 1 mol of KHCO3
No. of moles of KHCO3 used
Show calculations below
No. of moles of HCl added
Show calculations below:
Assuming KHCO3 is the limiting reactant, how many grams of KCl can be formed?
Show calculations below:
Assuming HCl is the limitingʻreactant, how many grams of KCl can be formed?
Show calculations below:
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Which reactant is excess?
Actual yield of KCI:
Theoretical yield of KCl:
Percent yield of KCl:
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IV.
Data Collection
Original NaOH Stock solution concentration (M): 10.0 NM
Volume of stock used (mL): 20.000ML
Total volume of solution in flask after dilution (mL): 100.000ML
New NaOH solution concentration (Working Stock) (M): 2.0M
Hint: Label out what each molarity and volume above is, and then calculate the New NAOH solution
concentration using M1V1 = M2V2
Table 1. Raw titration data
Trial 1
50.000 M L
24.500mし
23.500ML
Trial 2
Trial 3
33.000ML
9,500ML
Initial Buret Reading
Final Buret Reading
Volume of NaOH (aq) used
Molarity of NaOH (aq) used
Volume of HCI (aq) used
Color at equivalence point
24.500ML
3.000 ml
23.500ML
23.500 mL
20.0 ML
20.0 ML
Pink
20.0ML
pink
pink
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Trial 1
Trial 2
Trial 3
Moles of NaOH used in
titration
Moles of HCI neutralized in
unknown sample
Molarity of HC
Average Molarity
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= 2.0M
(10.0M) (20.000 g)= uk (100. 00ome)=
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100 A
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Maps
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dlo
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at concentration should the student write down in her lab notebook? Be sure your answer has the correct number of significant digits.
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<
STARTING AMOUNT
X
A brine solution is 3.50% NaCl by mass. Given that its density is 1.071 g/mL
, determine the quantity of liters of solution that contains 5.00 moles of NaCl
5.00
ADD FACTOR
0.1
*( )
3.50
10
35.45
Question 29 of 33
1.071
0.01
8.03 x 1022
1000
100
ANSWER
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22.99
0.0156
L solution g NaCl/mL g solution mL solution
kg solution
g NaCl/g solution
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moles of NaCl
mol NaCl
RESET
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2
1
g NaCl
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Sample 1
Sample 2
Sample 3
Mass of Erlenmeyer Flask (g)
24.33
24.37
24.44
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(g)(lime water)
27.22
27.29
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2.89
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2.89
2.92
2.89
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0.1
0.1
0.1
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1.00+1.00
1.00+1.00
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.66
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1.34
1.46
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Molar Solubility (M)
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Sample 2
Sample 3
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24.37
24.44
Mass of Erlenmeyer Flask + Calcium Hydroxide Solution (g)
(g)(lime water)
27.22
27.29
27.33
Mass of Calcium Hydroxide Solution (g)
2.89
2.92
2.89
Volume of Ca(OH)2
Density = 1.000 g/mL
2.89
2.92
2.89
Concentration of HCl (M)
0.1
0.1
0.1
Initial HCl Volume in Syringe (mL)
1.00+1.00
1.00+1.00
1.00+1.00
Final HCl Volume in Syringe (mL)
.72
.66
.54
Volume of HCl Delivered (mL)
1.28
1.34
1.46
Moles of HCl Delivered
Moles of OH- in Sample
Moles of Ca2+ in Sample
Molar Solubility (M)
Calculated Ksp
Average Calculated Ksp
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