Molar mass of Al₂(CrO₄)₃ is the mass of one mole of Al₂(CrO₄)₃
one mol =weight of Al₂(CrO₄)₃ in g /one mol
1) We have to find the atomic mass (in a periodic talbe) of the following elements (in u)
Al=26.98 u
Cr=52 u
O=16 u
2)We calculate the atomic weight of Al₂(CrO₄)₃
Atomic weight (Al₂(CrO₄)₃)=2(26.98 u)+3[(52 u)+4(16 u)]=
=53.96 u+3(52 u+64 u)=53.96 u+3(116 u=53.96 u+348 u=401.96 u
3) One mole of (Al₂(CrO₄)₃ ) have 401.96 g of Al₂(CrO₄)₃
molar mass=401.96 g/ mol
answer: 401.96 g/ mol
All of the acid molecules in beaker 1 dissociate fully and exist as and ions. As a result, beaker 1 represents a strong acid solution. The majority of the molecules in beaker 2 are undissociated.
Na - 1s^2, 2s^2, 2p^6, 3s^1
Ne - 1s^2, 2s^2, 2p^6
O - 1s^2, 2s^2, 2p^4
The axial positions the bond angle is 120 degrees and in equatorial positions the bond angle is 90 degrees.
Functional groups on central atom gets reduce if lone pairs are added.
Explanation:
The number of lone pairs and base pairs of electrons tells the geometry of the molecule.
VSEPR Theory helps to know the lone pairs and bond pair electrons on the centre atom of the molecule.
Example of molecule containing 5 electron pairs can have four bond pairs and 1 lone pair.
eg: Cl
the repulsion is shown as
lp-lp> lp-bp>bp-bp
These are in equatorial position because of the repulsion of lone pairs.
It can have 2 lone pairs and 3 bond pairs. eg. Xe
Lone pairs in this is also in equatorial position as
lp-lp> lp-bp> bp-bp
In axial positions the bond angle is 120 degrees
in equatorial positions the bond angle is 90 degrees, due to the repulsion in lone pair of electrons.
If 1 lone pair is there it can be replaced by bonding with hydrogen.
If 2 lone pairs are there then bonding with oxygen is there. The covalent bond is formed.
Answer:
1.33 M
Explanation:
We'll begin by writing out the data obtained from the question. This includes the following:
Volume of the stock solution (V1) = 0.5L
Molarity of the stock solution (M1) = 4M
Volume of diluted solution (V2) = 1.5L
Molarity of the diluted solution (M2) =.?
With the application of the dilution formula, the molarity of the diluted solution can be obtained as follow:
M1V1 = M2V2
4 x 0.5 = M2 x 1.5
Divide both side by 1.5
M2 = (4 x 0.5) / 1.5
M2 = 1.33 M
Therefore the molarity of the diluted solution is 1.33 M
