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Question 3.15: Calculate the density of UO2?

i keep getting 16.85 g/cm^3, but the answer in the back of thebook is 10.9 g/cm^3.

I dont's understand what i am doing wrong.

a=0.474nm

Density = m/V

= ([4(238.030)+8(16.0)]/0.6023*10^24 g /(0.474nm)^3 * (10^7nm/1cm)^3

= 16.85 g/cm^3

What i'm i doing wrong? please help.

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Jarrod Robel
Jarrod RobelLv2
28 Sep 2019

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Related questions


I am trying to do this problem but am getting a differentanswer. A) I got 0.014 mL(same) B) I got 7x10^-5 (not 6.85 am Idoing something wrong?) C)1100cm^2 (this is where I get lost) Cansome explain these a little better. Thanks

Given Data: average number of drops/mL = 73.0; overall averagediameter of oleic acid film = 38.0 cm. Use this data to calculatefollowing:

a) average volume of oleic acid in film
b) volume of oleic acid per drop of solution
c) average area of oleic acid film
d) height of oleic acid film
e) volume of one oleic acid molecule
f) Avogadro's number
g) Percent disagreement

Formulas to be used:
1) Average volume of one drop of oleic acid solution:
V1 = 1.00 mL / average # of drops
2) Average volume of oleic acid in film in cm^3:
V2 = V1 x (0.0050 mL / 1 mL) x (1 cm^3 / 1 mL)
3) Height of oleic film in cm:
H = V2 in cm^3 / Average area of film in cm^3
* 1 mL = 1 cm^3
4) Volume of one oleic acid molecule in cm^3. A cubic shape isassumed:
V3 = H^3
5) Avogadro's number:
N = (282 g/mol) / (V3 x 0.895 g/cm^3)
6) Percent disagreement:
% dsgmt = [log(6.0225 x 10^23) - log(N)] / log(6.0224 x 10^23) x100


I know how to plug in the values with the equations. However, Ikeep getting a 100%+ disagreement which is unacceptable. I wouldlike some help please. Here are the results that I amgetting:
a) 6.8 x 10^-5 cm^3
b) 1.4 x 10^-2 mL
c) 1134 cm^2
d) 6.04 x 10^-8 cm
e) 2.203 x 10^-22 cm^3
f) 1.43 x 10^24
g) 100%+ error.

Before anyone deletes this question, I am aware that it is not"savvy" to ask more than one question in this question, but thereason I did is that they all relate to one another and cannot besolved without solving the others.

Verifying the moles and mass of C6O6H8 from KIO3 titration and determing average % of vitamin C.

Experiment: Measured 50.0 mL of apple juice in an erlenmeyer flask. Add 1g KI and 5mL of 1 M HCI to the mixture. Add 2-3 mL of 0.5% starch solution. Titrate with 0.0100 M KIO3 through a burrette. Repeat 3 times.

Results:

Juice 1 Juice 2 Juice 3
Juice Volume (mL) 50.0 50.0 50.0
mL of KIO3dispensed (mL) 5.15 5.92 4.98
moles of IO3-
moles of C6O6H8
Mass of C6O6H8 (mg)
Male % Vitamin C
Female % Vitamin C

What I have tried:

To find the number of IO3- moles:

[(mL of KIO3 dispensed) * (0.001)] * (0.0100 M KIO3) = moles IO3-

To find the number of C6O6H8 moles:

[(mL of KIO3 dispensed) * (0.001)] * (0.0100 M KIO3) *3 = moles C6O6H8

To find the mass of C6O6H8:

([(mL of KIO3 dispensed) * (0.001)] * (0.0100 M KIO3) *3) * 176.124 mol C6O6H8 = mass of C6O6H8 (then convert to mg)

When I do this I get the following masses for C6O6H8:

Juice 1 Juice 2 Juice 3
27.057 mg 31.280 mg 26.312 mg

The issue I'm having when calculating the mass of vitamin C in a single serving (240 mL) for each titration.

I'm assuming since 240/50 = 4.8, you would multiply the mass of C6O6H8 by 4.8 for each titration. But I'm getting really big numbers when doing this. And the reason I'm would like help or clarification is because the next question states:

"The % Vitamin C is based on the recommended daily allowance which is 75 mg for females and 90 mg for males. Calculate the % of vitamin C based on your results from the mass of vitamin in a single serving (240mL) for the average male and average female"

But if 240 mL has 120% of the daily recommended allowance of Vitamin C, then 240 mL of apple juice should contain between 90 mg to 108 mg of Vitamin C which is much less than the results I am getting. I am not sure if I am reading the question wrong or if my calculations are wrong. Any help and explanations would be appreciated. Thank you!

I worked question #3 and I got 463576 (g/mol)

I used Beers law ( A= Ecl) to calculate the concentration (M) of the solution. I then used the volume to give me moles for 75 uL of sample.

I then used the density to give me grams.

Once I had grams and moles I used these pieces to calculate molar mass.

I feel like I solved this corretly, but I didn't use the buffer in my calculations and I was wondering if it would have an effect. If you have a different way please solve and show your work.

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