Answer:
11445.8years
Explanation:
Half-life of carbon-14 = 5720 years
First we have to calculate the rate constant, we use the formula :
The balanced equation is 2
AlI
3
(
a
q
)
+
3
Cl
2
(
g
)
→
2
AlCl
3
(
a
q
)
+
3
I
2
(
g
)
.
<u>Explanation:</u>
- Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- .
Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.
- Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.
Balancing the equation, we get:
2AlI
3( aq ) + 3Cl2
( g ) → 2AlCl3
( aq )
+ 3
I
2 ( g )
-
Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.
-
Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.
-
We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
Answer:
Potassium selenide
Explanation:
Potassium selenide (K2Se)
Answer:
The answer is quartet 2.40 ppm.
Note: Kindly find an attached image below for the part of the solution to this question
Sources: The image was researched from Course hero platform
Explanation:
Solution
Multiplicity or (n+1) rule:
It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.
Multiplicity =(n+1)
Thus
The non equivalent protons which are attached from adjacent atom is denoted by n.
Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:
The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:
Non- equivalent = 3
Multiplicity = (3 +1)
= 4
= Quartet for 2H
A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.
Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.
Although there isn’t a picture a graph can be misleading when it doesn’t start at zero, it doesn’t give accurate information, it skips too many numbers, the vertical scale is too big or too small. Hope this helps
