What volume of ar at 53.6 °c and 1.31 atm contains the same number of particles as 0.584 l of h2 at 19.6 °c and 2.10 atm?

Volume = 1.043L

Explanation:

First off, we have to calculate the number of particles H2 at that condition have.

We were given the following;

Volume (V) = 0.584L

Temperature = 19.6°C + 273 = 292.6 K (Converting to Kelvin Temperature)

Pressure = 2.10 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making n subject of formular;

n = PV / RT

n = (2.1 * 0.584) / (0.0821 * 292.6)

n = 1.2264 / 24.02246

n = 0.05098 moles

Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×1023 mol-1

Number of particles present in 0.05098 moles of H2 is given as;

1 mole = 6.022×1023

0.05098 = x

Upon cross multiplication, we have;

x = 0.05098 * 6.022×1023 = 0.307×1023 = 3.07×1022 particles.

equal number of moles would have the same number of particles. So we just have to find the volume when the number of moles of Ar is 0.05098.

We were given the following for Ar;

Volume (V) = ?

Temperature = 53.6°C + 273 = 326.6 K (Converting to Kelvin Temperature)

Pressure = 1.31 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making V subject of formular;

V = nRT / P

V = ( 0.05098 * 0.0821 * 326.6) /  1.31

V = 1.043L

the movement of aircraft through clouds can even cause lightning strikes. a single lightning event is a "flash", which is a complex, multi-stage explosively ejected.

explanation:

and the person below goggled it.

mark as brainliest

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.