Considering ideal gas:
PV= RTn
T= 25.2°C = 298.2 K
P1= 637 torr = 0.8382 atm
V1= 536 mL = 0.536 L
:. R=0.082 atm.L/K.mol
:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))
:. n= O.0184 mol
Then,
P2= 712 torr = 0.936842 atm
V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)
:.V2 = 0.4796 L
OR
V2 = 479.6 ml
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Answer:
I think it's 6 moles are produced
False because particles stop moving or study slow down.
Answer:
you can see the answer at the pic
