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:
lighting a match is the answer.
The orbitals closest to the nucleus is the orbital wih the lowest energy. This is according to the basic rules stating that the energy of the shells as its principal quantum number increases, also increases. Thus the answer in 1 is B. Valence electrons are found in the outermost electron shell, on the other hand.
Answer: Potassium Chloride
Explanation:
Answer:
Part A: 47.8 mi/h
Part B: 0.072 M/s
Part C: 0.144 M/s
Explanation:
Part A
The average speed or velocity (V) is the variation of the space divided by the variation of the time:
V = (241 - 2)/(8 -3)
V = 47.8 mi/h
Part B
As Part A, the average rate (r) of formation of I2 is the variation of the concentration divided by the variation of time:
r = (1.83 - 1.11)/(15 - 5)
r = 0.072 M/s
Part C
The rates of the substances are proportional of their number of moles (n) which are their coefficient, so:
rI2/nI2 = rHCl/nHCl
0.072/1 = rHCl/2
rHCl = 2*0.072
rHCl = 0.144 M/s
