Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

CHEM 103 – Portage Learning – Module 6 Exam Questions and Verified Answers (2024/2026) – C, Exams of Nursing

CHEM 103 – Portage Learning – Module 6 Exam Questions and Verified Answers (2024/2026) – Complete Test Preparation Material This document includes all the verified questions and answers for Module 6 of CHEM 103 from Portage Learning. It covers essential concepts related to acids and bases, pH calculations, neutralization reactions, and buffer systems. The content is aligned with the module objectives and serves as a reliable resource for achieving a high score on the exam.

Typology: Exams

2024/2025

Available from 05/23/2025

Nurseexcel
Nurseexcel 🇺🇸

5

(1)

832 documents

1 / 8

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Question 1
Not yet graded / 10 pts
Question 2
Not yet graded / 10 pts
Question 3
Not yet graded / 10 pts
CHEM 103 Module 6 Exam Portage Learning
Questions and Verified Answers,100% Guaranteed Pass
CHEM 103
MODULE 6 2025/2026
Click this link to access the
Periodic Table.
This may be helpful throughout
the
exam.
List and explain if each of the following solutions conducts an electric current:
sodium chloride (NaCl), hydrochloric acid (HCl) and sugar (C
6
H
12
O
6
).
Sodium chloride (NaCl) is an ionic compound and conducts since it forms ions in
solution. Hydrochloric acid (HCl) is a polar compound and conducts since it
forms ions in solution. Sugar (C
6
H
12
O
6
) is a molecular compound but does not form
ions in solution so it does not conduct.
Explain how and why the presence of a solute affects the boiling point of a solvent.
The presence of a solute raises the boiling point of a solvent by lowering the vapor
pressure of the solvent. With this lower vapor pressure, more heat (a higher
boiling point) is required to raise the vapor pressure to atmospheric pressure.
Click this link to access the
Periodic Table.
This may be helpful throughout
the
exam.
pf3
pf4
pf5
pf8

Partial preview of the text

Download CHEM 103 – Portage Learning – Module 6 Exam Questions and Verified Answers (2024/2026) – C and more Exams Nursing in PDF only on Docsity!

Question 1

Not yet graded / 10 pts

Question 2

Not yet graded / 10 pts

Question 3

Not yet graded / 10 pts

CHEM 103 Module 6 Exam Portage Learning

Questions and Verified Answers,100% Guaranteed Pass

CHEM 103

MODULE 6 202 5/

Click this link to access the Periodic Table. This may be helpful throughout the exam. List and explain if each of the following solutions conducts an electric current: sodium chloride (NaCl), hydrochloric acid (HCl) and sugar (C 6 H 12 O 6 ). Sodium chloride (NaCl) is an ionic compound and conducts since it forms ions in solution. Hydrochloric acid (HCl) is a polar compound and conducts since it forms ions in solution. Sugar (C 6 H 12 O 6 ) is a molecular compound but does not (^) form ions in solution so it does not conduct. Explain how and why the presence of a solute affects the boiling point of a solvent. The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure. Click this link to access the Periodic Table. This may be helpful throughout the exam.

Question 4

Not yet graded / 10 pts

Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare, List from lowest freezing point to highest freezing point. GaCl 3 , Al 2 (SO 4 ) 3 , NaI, MgCl 2 Your Answer: Al2(SO4)3 - > 2Al3+, 3SO4 -^2 = = 5 ions - > most ions = lowest freezing point GaCl2 - > Ga3+, Cl-^ = 4 ions MgCl2 = 3 ions NaI = 2 ions - > least ions = highest freezing point GaCl 3 3rd lowest FP → (^) Ga+3 + 3 Cl-^ ∆tf = 1.86 x 0.1 x 4 = Al 2 (SO 4 ) 3 lowest FP → 2 Al+3^ +

3 SO 4

  • 3 ∆tf^ =^ 1.86^ x^ 0.1^ x^5 = NaI = highest FP → Na+^ + I-^ ∆tf = 1.86 x 0.1 x 2 MgCl 2 = 2nd lowest FP → Mg+2^ + 2 Cl-^ ∆tf = 1.86 x 0.1 x 3 FP: Al 2 (SO 4 ) 3 < GaCl 3 < MgCl 2 < NaI Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the mass percent solute in a solution of 20.8 grams of Ba(NO 3 ) 2 in 400 grams of water. Report your answer to 3 significant (^) figures.

exam.

Question 8

Not yet graded / 10 pts

Show the calculation of the mass of Ba(NO 3 ) 2 needed to make 250 ml of a 0.200 M solution. Report your answer to 3 significant figures. Molarity = (moles) / (mlsolvent / 1000) 0.200 = (moles) / (250 / 1000) Moles = 0.200 x 0.250 = 0. Moles = (gsolute / MW) 0.0500 = (gsolute / 261.55) gsolute = 0.0500 x 261.55 = 13.1 g Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NO 3 ) 2. molessolute = gsolute / MW molessolute = 37.5 g / 261.55 = 0.1434 mol Molarity = moles / (mL /1000)

Question 9

Not yet graded / 10 pts

Question 10

Not yet graded / 10 pts

mL / 1000 = 0.1434 / 0.667 = 0. mL = 0.2150 x 1000 = 215 mL Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the boiling point of a solution made by dissolving 20.9 grams of the nonelectrolyte C 4 H 8 O 4 in 250 grams of water. Kb for water is (^) 0.51, BP of pure water is 100oC. Calculate your answer to 0.01oC. molality = (gsolute / MW) / (gsolvent / 1000) molality = (20.9 / 120.104) / (250 / 1000) = 0.6961 m ∆tb = Kb x m = 0.51 x 0.6961 = 0.355oC BPsolution = BPsolvent - ∆tb = 100 oC + 0.355 = 100.35oC Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molar mass (molecular weight) of a solute if a solution of 14.5 grams of the solute in 200 grams of water has a freezing point^ of^ - 1.35oC. Kf for water is 1.86 and the freezing point of pure water is 0 oC. Calculate your answer to 0.1 g/mole.

∆tf = Kf x m molality = ∆tf / Kf = 1.35 / 1.86 = 0.726 m molality = (gsolute / MW) / (gsolvent / 1000) 0.726 = (moles) / (200 / 1000) Moles = 0.726 x 0.200 = 0. 0.1452 = (14.5 / MW) MW = 14.5 / 0.1452 = 99.