Answer:
A. 1:3
Explanation:
If we look at the ions shown in the image attached to the question, we will notice that we have aluminum (Al^3+), a trivalent ion combining with the iodide ion (I^-).
Aluminum can easily give out its three outermost electrons to three atoms of iodine. If aluminum gives out its three electrons, it achieves the stable octet structure. Iodine atoms have seven electrons in their outermost shell. They only need one more electrons to complete their octet. This one electron can be gotten by the combination of three iodine atoms with one atom of aluminum. One electron each is transferred from the aluminum atom to each iodine atom to form AlI3 with a ratio of 1:3.
Answer: -
6
Explanation: -
The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2
We see there 3 sodium on the right side from Na3AsO3.
But there are only 1 sodium on the left from NaOH.
So we multiply NaOH by 3.
As + 3 NaOH -- > Na3AsO3 + H2
Now we see the number of Hydrogen on the left is 3.
But the number of hydrogens is 2 on the left.
So, we multiply to get both sides 6 hydrogen.
As + 6NaOH -- > Na3AsO3 + 3 H2
Rebalancing for Na,
As + 6NaOH -- > 2Na3AsO3 + 3 H2.
Finally balancing As,
2 As + 6 NaOH -- > 2Na3AsO3 + 3H2
The coefficient of the NaOH molecule in the balanced reaction is thus 6
Answer: The molecular mass of this compound is 131 g/mol
Explanation:
Depression in freezing point:
where,
= depression in freezing point = 
= freezing point constant = 
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute = 0.49 g
= mass of solvent (cyclohexane) = 20.00 g
= molar mass of solute = ?
Now put all the given values in the above formula, we get:
Therefore, the molar mass of solute is 131 g/mol
Answer:
Empirical CHO
molecular C4H4O4
Explanation:
From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.
This means : O = 100 - 41.39 - 3.47 = 55.14%
Next is to divide the percentage compositions by their atomic masses.
C = 41.39/12 = 3.45
O = 55.14/16 = 3.45
H = 3.47/1 = 3.47
Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.
Hence the empirical formula of Maleic acid is CHO
Now we go on to deduce the molecular formula.
To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.
Now we can see that 0.378mole = 43.8g
Then 1 mole = xg
x = (43.8*1)/0.378 = 115.87 = apprx 116
[CHO]n = 116
(12 + 1 + 16]n = 116
29n = 116
n = 116/29 = 4
The molecular formula is thus C4H4O4
<span>There is a direct correlation between the period number and the energy level for valence electrons. For example, the H and He elements, in period 1, have their outer electrons in the energy level "1". This continues down the rows: all the elements in period 2 have their principal energy level as n = 2, period 3 has n = 3, and so on.</span>
