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CHEM 120 Final Exam (Version 3), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing. CA$22.48   Add to cart
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CHEM 120 Final Exam (Version 3), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.
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CHEM 120 Final Exam (Version 3), CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.

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  • May 3, 2023
  • 25
  • 2022/2023
  • Exam (elaborations)
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  • chem 120 final exam version 3
  • chem 120 introduction to general
  • organic amp biological chemistry with lab
  • verified and correct answers
  • chamberlain college of nursing

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  • CHEM 120, Chamberlain college of Nursing
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CHEM 120 Week 8 Final Exam
Course Number: CHEM120
Course Title: Introduction to General,Organic & Biological Chemistry with Lab


 Correct Question & Answers
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,CHEM 120: Final Exam


Multiple Choice
1. What is the molarity of a solution containing 55.8 g of MgCl2 dissolved in
1.00 L of solution?
M= moles of solute
liters of solution.
We are given the mass of our solute so we just need to convert
it to moles of MgCl2 via the MM.
55.9 g MgCl2 x 1 mol MgCl2 =0.586111 mols MgCl2
95.2119 g MgCl2
Now we can plug this into our equation for molarity
M = 0.587111 mols MgCl2 = 0.586 M MgCl2
`1.00 L
a) 55.8 M
b) 1.71 M
c) 0.586 M
d) 0.558 M

2. What is the mass in grams of Mg(NO3)2 present in 145 mL of a 0.150 M
solution of Mg(NO3)2?
We know the molarity and the volume, but we need to find the moles.
M= moles of solute
liters of solution. (volume)

First convert mLL.
145 mL 0.145 L (divide by 1000)
so we can solve this equation for moles.
Moles= (M) x (Volume)
moles Mg(NO3)2 = (0.150 moles) x (0.145 liters) = 0.02175 moles
Mg(NO3)2
liter

Now we can convert the number of moles to grams via the molar
mass of Mg(NO3)2.

0.02175 moles Mg(NO3)2 x 148.3 grams Mg(NO3)2 = 3.225 grams
Mg(NO3)2
1 mol Mg(NO3)2

, a) 3.23 g
b) 0.022 g
c) 1.88 g
d) 143 g

3. When aqueous solutions of __________ are mixed, a precipitate forms.
To work out this problem, write molecular equations for each
reaction and, using your solubility chart, look at the products to
see if a precipitate is formed.
a) NiBr2 and AgNO3 : NiBr2(aq) + AgNO3(aq)  NiNO3(aq) + 2
AgBr (s)
Nitrates are always soluble. Br- is usually soluble but when
paired with Ag, it forms a precipitate.
b) NaI and KBr- NaI (aq) + KBr(aq)  NaBr (aq) + KI (aq)
Both of the products for this reaction are aqueous so no
reaction will occur.
c) K2SO4 and CrCl3 – 3 K2SO4 (aq) + 2 CrCl3(aq)  6 KCl (aq) +
Cr2(SO4)3 (aq)
Chlorides and sulfates are generally soluble and they both
form aqueous products so there is no reaction.
d) KOH and Sr(NO3)2 – KOH (aq) + Sr(NO3)2 (aq)  KNO3(aq) +
Sr(OH)2 (aq)
Even though OH- groups are usually insoluble, Sr2+ is one of
the exceptions.
) Li2CO3 and CsI - Li2CO3 (aq) + CsI (aq)  LiI(aq) + Cs2CO3 (aq)
Iodides
Carbonates are soluble when paired with alkali metals and so
are iodides. No reaction.

4. The net ionic equation for the formation of aqueous copper (II)
perchlorate which occurs when an aqueous solution of perchloric acid is
mixed with solid copper hydroxide is given by:
Perchloric acid: HClO4
Copper hydroxide: Cu(OH)2
Copper (II) perchlorate has the formula: Cu(ClO4)2
First write the balanced molecular equation:
2 HClO4 (aq) + Cu(OH)2 (s)  Cu(ClO4)2 (aq) + 2 H2O (l)
Then write the ionic equation (breaking up the aqueous
reactants/products)

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