The answer to this is a solution
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
Ight even though you have to answer other peeps
The rule is number your table 1 2. 3 4 5 6 7 0 the number of shells increase as you go further up as it indicates the number of electrons on the outer shell e.g Argon will have 3 electrons on its outer shell hope this helps if not its then tighttt
Answer:
Piso = 32.17 Torr
Pprop = 5.079 Torr
yprop = 0.1364
yiso = 0.8636
Explanation:
From the question; we can opine that :
- The mole fraction of isopropanol in a mixture of isopropanol and propanol will be 1.
-
The partial pressure of isopropanol will be its mole fraction multiplied by vapor pressure of isopropanol
-
The partial pressure of propanol will be its mole fraction multiplied by vapor pressure of propanol
-
In the vapor, the mole fraction of propanol will be its partial pressure divided by the sum of the two partial pressures
NOW;
When xprop = 0.243; xisopropanol will be 1- 0.243 = 0.757
P°iso = 45.2 Torr at 25 °C so
Piso will be 45.2 × 0.757 = 32.17 Torr
Pprop will be 20.9 × 0.243 = 5.079 Torr
yprop = 5.079/(5.079 +32.17) = 0.1364
yiso = 1-0.1364 = 0.8636
