dentistguinness

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Could you help me solve these questions?

1. If 3.00 g of a nitrogen-oxygen compound is found to contain 2.22 g of oxygen, what is the percentage of nitrogen in the compound?
a. (3.00/2.22)(100)
b. [(3.00 + 2.22)/3.00](100)
c. (2.22/3.00)(100)
d.[(3.00-2.22)/3.00](100)
e. [(3.00-2.22)/3.00](100)

2. A 10.0 liter sample of oxygen at 100 degrees Celsius and 1 atm is cooled at 27 degrees Celsius and expanded until the pressure is 0.5 atm. Find the volume of the oxygen.

3. How many grams of NaOH (40 g * mol^-1) are there in 250 mL of 0.4 M NaOH solution?

4. Which will be the final volume when 400 mL of 0.6 M HCl is diluted to 0.5 M HCl?
a. 400(0.5/0.6)
b. (0.6-0.5)(400)
c. (400)(0.6/0.5)
d. (1,000-400)(0.5/0.6)
e. (0.6/0.5)(1,000-400)

5. If 25 mL of 0.5 M NaOH neutralizes 35 mL of a monoprotic acid, which is the molarity of the acid?
a. [(0.025)(0.5)]/0.035
b. 0.065/[(0.025)/(0.5)]
c. 0.025/[(0.5)(0.035)]
d. (0.025)(0.5)(0.035)
e. [(0.035)(0.5)]/0.025
 
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artist2022

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I have a question- what's the difference in answers d & e of question 1?

1. You're given total weight, and the weight of Oxygen. You need to find Nitrogen: (total - oxygen) = nitrogen amount
percent nitrogen: [(total-oxygen)/total] * 100 ---> [(3.00-2.22)/(3.00)]*100 = answer d or e? (they look the same? lol)

2. Ideal gas law, with variation: Vf/Vi = (nRTf/Pf)/(nRTi/Pi) (where f = final, and i = initial)
so, we need to solve for Vf ---> Vf = [(nRTf/Pf)/(nRTi/Pi)] * Vi
n and R stay constant, they cancel out (bolded) ---> Vf = [(Tf/Pf)/(Ti/Pi)] * Vi
plug in your numbers given, make sure T is in Kelvins ----> Vf = [(300/0.5)/(373/1)] * 10 = 16.08 L (makes sense, because the pressure decreased which means the volume must increase)
 

dentistguinness

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Jun 17, 2017
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I have a question- what's the difference in answers d & e of question 1?

1. You're given total weight, and the weight of Oxygen. You need to find Nitrogen: (total - oxygen) = nitrogen amount
percent nitrogen: [(total-oxygen)/total] * 100 ---> [(3.00-2.22)/(3.00)]*100 = answer d or e? (they look the same? lol)

2. Ideal gas law, with variation: Vf/Vi = (nRTf/Pf)/(nRTi/Pi) (where f = final, and i = initial)
so, we need to solve for Vf ---> Vf = [(nRTf/Pf)/(nRTi/Pi)] * Vi
n and R stay constant, they cancel out (bolded) ---> Vf = [(Tf/Pf)/(Ti/Pi)] * Vi
plug in your numbers given, make sure T is in Kelvins ----> Vf = [(300/0.5)/(373/1)] * 10 = 16.08 L (makes sense, because the pressure decreased which means the volume must increase)
Ah I put down the wrong number for E. It is D though! Choice E was suppose to say 2.22 not 3.00, thanks for catching that.
What does the nRTf and Pf stand for in question 2?
 

artist2022

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Ah I put down the wrong number for E. It is D though! Choice E was suppose to say 2.22 not 3.00, thanks for catching that.
What does the nRTf and Pf stand for in question 2?
So I put an "f" for "final"
So, it was (Temperature Final/Pressure Final)/(Temperature Initial/Pressure Initial). You were given the initial and final amounts for some variables, and this is how you incorporate them. You were also given the Initial Volume, and had to find Final Volume.
n represents number of moles, which stayed the same so it canceled out.
R represents the gas constant, which is the same so it canceled out.
I'll do the remaining three too since no one else answered

3. 0.4M NaOH = 0.4 mol/L NaOH
Given 0.250 L of solution
0.4 mol/L * 0.250L --> 0.1 mol NaOH
Use its molar mass to calculate grams:
0.1 mol NaOH (40g/mol) ---> 4 g NaOH

4. Use this equation: m1v1=m2v2
(0.6M)(400mL) = (0.5M)(x mL)
x = [(0.6)(400)]/(0.5) ---> answer is C

5. If it neutralizes, that means their moles were equivalent to each other.
Molarity is moles/volume.
How many moles of NaOH were there? 0.5 M x 0.025 L = # moles (we don't care about the number)
I said earlier that the moles were equivalent, so how many moles of the acid were there? # moles acid = # moles NaOH = (substitute!) 0.5*0.025
How much acid were we given? 0.035L
So to find its molarity, it will be moles acid/volume acid
(0.5*0.025)/(0.035) ---> answer is A

Hopefully this helps!
 
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