The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
We are given the number of moles:
O = 1.36 mol
H = 4.10 mol
C = 2.05 mol
To get the empirical formula, first divide everything by
the smallest number of moles = 1.36 mol. So that:
O = 1 mol
H = 3 mol
C = 1.5 mol
Next step is to multiply everything by a number such that
all will be a whole number. In this case, multiply by 2 to get a whole number
for C, so that:
O = 2 mol
H = 6 mol
C = 3 mol
Therefore the empirical formula of the compound is:
C3H6O2
Answer:
See below
Explanation:
The glycosidic bond forms between the carbon atom on the glucose molecule and the nitrogen atom present on the nitrogenous base. A diagram has been attached to show this particular bond. There is also a phosphate molecule bonded on the sugar molecule at the other end.
Answer:
the oxygenmolecule(02or0-0) its self is free redicles or a-b redical whith two unpaired electrons that is electron configuration of the molecule has two unpaired electros occupying two degenerate molecular or bitals with the same spin orientation(up-upordown-dowan
Explanation:
in the lower atomsphere important redical are produced by the photodissociation if nitrogen dioxide to an oxygen atom and nitric oxide