The pressure of an ideal gas is 782mmHg at 35°C in a box with a volume of 2,239mL. What is the number of moles? If this gas is oxygen, O2, what is the number of molecules present?

The vessel has 0.0913 moles of O₂. Therefore it has 5.49ₓ10²² molecules of O₂

Explanation:

Let's apply the Ideal Gases Law to solve this.

First of all, we convert the pressure from mmHg to atm

782 mmHg . 1 atm / 760 mmHg = 1.03 stm

We convert the volume from mL to L

2239 mL . 1L / 1000mL = 2.239L

and finally we convert T° C to K → 35°C + 273 = 308K

Formula is: P . V = n . R . T

1.03 atm . 2.239L = n . 0.082 L.atm/mol.K . 308K

n = 1.03 atm . 2.239L / 0.082 L.atm/mol.K . 308K → 0.0913 moles

Let's count the molecules  → 0.0913 mol . 6.02ₓ10²³ molecules / 1 mol =

5.49ₓ10²² molecules

9.1 * 10^-2 moles of gas

5.49*10^22 molecules O2

Explanation:

Step 1: Data given

Pressure of the gas = 782 mmHg = 782 / 760 atm  = 1.02895 atm

Temperature = 35 °C = 308 K

Volume = 2.239 L

Step 2: Calculate moles of gas

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles of gas = TO BE DETERMINED

⇒with p = the pressure of the gas = 1.03026 atm

⇒with V = the volume of the gas = 2.239 L

⇒with R = the gas constant = 0.08206 L*atm / mol *K

⇒with T = the temperature = 35 °C = 308 K

n = (p*V)/ (R*T)

n = (1.03026 * 2.239) / (0.08206*308)

n = 0.091268 moles of gas = 9.1 * 10^-2 moles of gas

Step 3: Calculate molecules O2

Number of molecules = moles * number of Avogadro

Number of molecules = 0.091268 moles * 6.02 * 10^23

Number of molecules = 5.49*10^22 molecules O2

There are 5.49*10^22 molecules O2

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1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.