100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
Previously searched by you
CHEM 120 Final Exam-Q & A (Version 2), Best document for preparation, Verified And Correct Answers Course Number: CHEM120 Course Title: Introduction to General, Organic & Biological Chemistry with Lab Chamberlain College of Nursing$15.49
Add to cart
CHEM 120 Final Exam-Q & A (Version 2), Best document for preparation, Verified And Correct Answers Course Number: CHEM120 Course Title: Introduction to General, Organic & Biological Chemistry with Lab Chamberlain College of Nursing
1 view 0 purchase
Course
CHEM 120
Institution
CHEM 120
CHEM 120 Final Exam-Q & A (Version 2), Best document for preparation, Verified And Correct Answers
Course Number: CHEM120
Course Title: Introduction to General, Organic & Biological Chemistry with Lab
Chamberlain College of Nursing
verified and correct answers course number chem120 course title introduction to general
organic amp biological chemistry
Written for
CHEM 120
All documents for this subject (558)
Seller
Follow
QuizPlus
Reviews received
Content preview
CHEM 120 Week 8 Final Exam
Course Number:CHEM120
Course Title:Introduction to General,Organic & Biological Chemistry with Lab
Correct Question & Answers
Secure high grade
Best feedback (rated) document by students
,CHEM: 120 Final Exam
1. (TCO 8) 35.0 mL of 0.25 M NaOH is neutralized by 23.6 mL of an HCl solution. The molarity of the HCl
solution is (show your work): (Points : 5)
2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5)
3.5 pints is equivalent to 1656.116
1 pint = 473.176 ml
3.5 pints* 473.176mL = 1656.116mL
0 1644446452 Short 6
3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5)
It's Zinc Phosphide
0 1644446453 Short 7
4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5)
Silver nitrate
0 1644446454 Short 11
5. (TCO 6) Calculate the pressure, in atmospheres, of 2.78 mol CO(g) in a 4.25 L tank at 51 degrees C.
(Points : 5)
Given that n = 2.78 mol; V = 4.25 L; and temperature
[2.78 mol* 0.0821 L -atm/ mol-K*304 K)/4.25 L = 16.3
6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant
, pressure, determine the volume at 25 degrees C. Show your work. (Points : 5)
Using Charles’ Law, (V1/T1) = (V2/T2).
First, convert temperature to KELVIN (T1 = t1 +273)
Thus, T1 = 95 + 273 = 368.
We have V1 (165 mL) & T2 = (25 + 273) = 298.
V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL.
0 1644446457 Short 16
7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant
temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5)
1021mL * 719 mm/745 mmHg = 985.36mL =985mL
Using Boyle’s law, P1V1 = P2V2. We have V1 (1021 mL), P1 (719 mmHg) and P2 (745
mmHg).
0 1644446459 Short 19
8. (TCO 12) If one strand of a DNA double helix has the sequence T T A G C G A C G C, what is the
sequence of the other DNA strand? (Points : 10)
AATCGCTGCG
0 1644446456 Short 14
1. (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of
grams of H3PO4 needed to prepare 0.75L of a 0.25M H3PO4 solution. Show your work.
(b, 5 pts) What volume, in Liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 2.5M
H3PO4 solution? Show your work. (Points : 10)
Using the molar mass given, convert this amount to grams.
mass = 0.1875 mol * (97.994 g/mol) = 18.37 grams H3PO4
b. M1*V1 = M2*V2
w here: M1 = 0.25; V1 = ??; M2 = 2.5; V2 = 50 mL = 0.050 L
Solvig for V1:
0.25 * V1 = 2.5 * 0.050
V1 = 0.50 L
2. First convert the given mass of NaOH to volume (in mL) using the density of NaOH
Volume = 43 g * (1 mL/2.13 g) = 20.19 mL
b. Volume % = volume of NaOH/ total volume
0.10 = 20.19 mL/total volume
Solving for total volume yields:
0 1644446463 Essay 3
2. (TCO 7) (a, 5 pts) What is the volume percent of a solution prepared by dissolving 21 g of NaOH in
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller QuizPlus. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $15.49. You're not tied to anything after your purchase.
