100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
Previously searched by you
CHEM120 Final Exam (4 Versions, Latest-2022)/ CHEM 120 Final Exam / CHEM120 Week 8 Final Exam / CHEM120 Unit 8 Final Exam: Chamberlain College of Nursing |Verified and 100% Correct Q & A|$45.49
Add to cart
CHEM120 Final Exam (4 Versions, Latest-2022)/ CHEM 120 Final Exam / CHEM120 Week 8 Final Exam / CHEM120 Unit 8 Final Exam: Chamberlain College of Nursing |Verified and 100% Correct Q & A|
1 view 0 purchase
Course
CHEM120
Institution
Chamberlain College Of Nursing
CHEM120 Final Exam (4 Versions, Latest-2022)/ CHEM 120 Final Exam / CHEM120 Week 8 Final Exam / CHEM120 Unit 8 Final Exam: Chamberlain College of Nursing |Verified and 100% Correct Q & A|
CHEM120 Week 8 Final Exam (4 Versions, Latest-2022)/ CHEM 120 Week 8 Final Exam / CHEM120 Final Exam / CHEM 12...
latest 2022 chem 120 final exam chem120 week 8 final exam chem120 unit 8 final exam chamberlain college of nursing |verified and 100 correct q amp a| chem120
Written for
Chamberlain College Of Nursing
CHEM120
All documents for this subject (99)
Seller
Follow
waldenquiz
Reviews received
Content preview
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): (Pts. : 5)
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).
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
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
3. (TCO 3) What is the name of the following compound: Zn3P2? (Points : 5)
, It's Zinc Phosphide
4. (TCO 3) What is the name of the following compound: AgNO3? (Points : 5)
Silver nitrate
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
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:
2. (TCO 7) (a, 5 pts) What is the volume percent of a solution prepared by dissolving 21 g of NaOH in
, enough water to make a final volume of 120 mL? Show your work.
(b, 5 pts) How many mL of a 10% solution can be made from the solution in part a? Show your work.
(Points : 10)
Part 1: based on the density of NAOH =2.1g/mL
then calculate the volume of 21g NaOHx(1mL/2.1g)=10 mL
so the percent volume is defined as
% volume = (Volume of Solute/volume of solution) x 100
so w e have % volume = (10mL/120mL)x100=8.3%
Part 2: you cannot prepare 10 % (volume) from the above solution (8.3%)
because the final solution is more concentrated than the initial
4. (TCO 11) Tungsten (W), with a mass number of 180 and an atomic number of 74, decays by emission
of an alpha particle. Identify the product of the nuclear reaction by providing its atomic symbol (5 pts),
mass number (5 pts), and atomic number (5 pts). (Points : 15)
mass of 180 becomes 176
atomic number of 74 becomes 72
name: Hafnium
Symbol: Hf
6. (TCO 13) What is the mRNA sequence for the following segment of DNA:
--AAACGTGTGCTAACA-- (10 pts)? Based upon the mRNA sequence, what is the peptide sequence (10
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 waldenquiz. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $45.49. You're not tied to anything after your purchase.
