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fomenos
3 years ago
15

Methyl alcohol can be used as a fuel instead of, or combined with, gasoline. A sample of methyl alcohol, CH3OH, in a flask at co

nstant volume exerts a pressure of 254 mm Hg at 57 degrees
c. The flask is slowly cooled.

a. assuming no condensation, use the ideal gas law to calculate the pressure of the vapor at 35 degrees
c.
Now I know using the first set of conditions I can find the moles and then do A that way, but I have a question first. When using the ideal gas law with the first set of conditions at constant volume, would I arbitrarily assign a volume to use in all calculations, or would I essentially leave it out of the ideal gas law?
Use an arbirtrary volume. Call it V, or if feel the urge, use 1 liter, that will but the moles on a per liter basis.
I have a follow-up question. I calculated the number of moles (.01233) and plugged it into the ideal gas equation to find the pressure at 35 degrees C and 45 degrees
c.312 atm and .322 atm, consecutively). These were the pressures of the vapor.
The next questions state:
b. Compare your answers in
a. with the equilibrium vapor pressures of methyl alcohol: 203 mmHg at 35 degrees C, and 325 mm Hg at 45 degrees
c.

c. based on your answers from
a. and
b., predict the pressure exerted by the methyl alcohol in the flask at 35 degrees C, at 45 degrees
c.

d. What physical states of methyl alcohol are present at both temperatures?
For b, I converted the mmHg pressures to atm (.2671 atm and .4276 atm consecutively), but I don't really get what the question is asking by "compare".
For c, I understand that I'm now supposed to find the pressure of the methyl alcohol, not the vapor as in
a. But what is the difference in finding this? How does
b. play a role in finding the pressure?
For d, is there an equation for this or should I just look it up?
Chemistry
1 answer:
andrew-mc [135]3 years ago
8 0
PV = nRT, or (pressure)(volume) = (amount of gas)(universal gas constant)(temperature) 

<span>You're solving for pressure, so start by rearranging the equation: </span>

<span>p = (nRT)/V </span>

<span>You're given temperature; all you need to do is convert it to Kelvin. (The universal gas constant is 8.31 J/(Kmol), so the units for temperature need to be the same.) </span>

<span>But how to figure out the rest? The question doesn't give you the size of the flask or number of moles, but it does tell you that these things don't change throughout the problem. Because they're constant, we don't have to know exactly what they are - we just have to know the *combined values* of the constants. (If this is confusing, think about it this way: a small amount of gas in a small flask will exert the same amount of pressure as a large amount of gas in a large flask, as long as the ratios are kept equal - it doesn't matter if it's 2 moles in 2 L, or 1 mole in 1 L.) </span>

<span>To figure out this constant, go back to the opening, where you're given temperature and pressure. </span>
<span>pV = nRT </span>
<span>Remember to convert celsius to kelvin! </span>
<span>(254mm Hg)(V) = nR(330K) </span>
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4 0
3 years ago
A 17.6-g sample of ammonium carbonate contains ________ mol of ammonium ions.
postnew [5]
These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate. 

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

17.6 gNH4CO3

Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse. 

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol


17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4

Now just take the molar mass we found to convert that amount into moles!

4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4

4 0
2 years ago
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit
Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0
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How many valence electrons does lithium (Li) have available for bonding?
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Lithium has 1 valence electron available for bonding. So its A.
6 0
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