Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?

The correct answer is option 2  (6 liters of N2)

Explanation:

The complete question

Which sample of gas at STP has the same number  of molecules as 6 liters of Cl2(g) at STP?

(1) 3 liters of O2(g)

(2) 6 liters of N2(g)

(3) 3 moles of O2(g)

(4) 6 moles of N2(g)

Step 1: Data given

Volume = 6 L

STP = 1 atm and 273 K

Step 2: Calculate moles of Cl2

p*V = n*R*T

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

⇒with n = the number of moles Cl2 = TO BE DETERMINED

⇒with V = the volume of Cl2 = 6.0 L

⇒with p = the pressure of Cl2 = 1 atm

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

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles Cl2

This means we need 0.2678 moles of a gas at STP

Option 3 has 3 moles of O2 ⇒ Not the same number of molecules

Option 4 has 6 moles of N2  ⇒ Not the same number of molecules

For 3 liters of O2 we'll have:

⇒with n = the number of moles O2 = TO BE DETERMINED

⇒with V = the volume of O2 = 3.0 L

⇒with p = the pressure of O2 = 1 atm

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

⇒ with T = the temperature = 273 K

n = (1.0 * 3.0 ) (0.08206 * 273)

n = 0,1339 moles ⇒ Not the same number of molecules

For 6 liters of N2 we'll have

⇒with n = the number of moles N2 = TO BE DETERMINED

⇒with V = the volume of N2 = 6.0 L

⇒with p = the pressure of N2 = 1 atm

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

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles N2 ⇒ The same number of molecules

The correct answer is option 2  (6 liters of N2)

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 : )

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.