Answer:
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
Explanation:
2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.
On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.
But ethanol reacts further with conc HI to give iodoethane.
Therefore,
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
This is all shown in the attached image.
Hope this Helps!!!
Answer:
Heres a picture of the periodic table.
Explanation:
Source(s):Science News for Students
Answer:
A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
Explanation:
Hi
The percentage of water in the sample is lower than expected.
A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:
If part of the sample is splashed from the crucible the mass of water detected will be less.
B. The dehydrated sample absorbed moisture after heating:
If the sample absorbs water after heating the percentage of water would be higher than expected.
C. The amount of the hydrate sample used was too small:
Depending on the sample size, different procedures can be chosen for analysis.
D. The crucible was not heated to constant mass before use:
In many occasions the crucible is heated next to the sample and not in previous form.
E. Excess heating caused the dehydration sample to decompose:
If the sample decomposes during heating, the analysis should be discarded.
success with your homework
Don’t currently have a calculator with me but just use Avogadros constant (A) 6.02x10^24, in the equation n=N/A. Lower case n being the number of mol and upper case being the number of molecules (given).
Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :

where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)

Therefore, the molal freezing point depression constant of X is 