Which sample would have the same number of molecules as 11.2L of He (g) at 273K and 202kPa?
1) 11.2L of N2(g) at 300K and 202kPa
2) 22.4L of Ne(g) at 546K and 404K
3) 11.2L of CH4(g) at 273K and 202kPa

11.2 L of at 273K and 202kPa

Explanation:

According to ideal gas equation:

P = pressure of gas = 202 kPa = 1.99 atm  ( 1kPa= 0.0098 atm)

V = Volume of gas = 11.2 L

n = number of moles = ?

R = gas constant =

T =temperature =

According to avogadro's law, equal number of moles occupy equal volumes and contain equal number of molecules at same temperature and pressure conditions.

As 11.2 L of at 273K and 202kPa will have same moles as 11.2L of He (g) at 273K and 202kPa, thus they have same number of molecules.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )