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
2Al +3CuSO4=Al2 (SO4)3+3Cu.
is the balanced equation.
HOPE IT HELPS YOU '_'
Answer:
14.93 g
Explanation:
First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:
- 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As
Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:
- 0.199 mol * 74.92 g/mol = 14.93 g
Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.
