Answer:
mNaCl = mNa + mCl
mNa = 20g
mCl = mCl2/2 = 30/2 = 15g
mNaCl= 20+15 = 35g
the reagents necessary to convert alcohol to ketone which involves oxidation of alcohols.
<h3>
What is oxidation of alcohols?</h3>
- Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
- In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
- In order to cause primary alcohols to oxidize into aldehydes
- (dichromate)
- /pyridine (Collins reagent)
- Chromium pyridinium compound (PCC)
- Dichromate of pyridinium (PDC, Cornforth reagent)
- Periodinane by Dess-Martin
- Oxalyl chloride with dimethylsulfoxide (DMSO) for Swern
- oxidation of secondary alcohols to ketones
- (dichromate)
- /pyridine (Collins reagent)
- Chromium pyridinium compound (PCC)
- Dichromate of pyridinium (PDC, Cornforth reagent)
- Periodinane by Dess-Martin
- Oxalyl chloride and dimethyl sulfoxide (DMSO) (Swern oxidation)
- /acetone (Jones oxidation)
- Acetone with aluminum isopropoxide (Oppenauer oxidation)
To learn more about oxidation of alcohols with the given link
brainly.com/question/7207863
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<u>Question:</u>
Identify the reagents necessary to achieve each of the following transformations
You will need the equation PV = nRT
P = Pressure in kPa
V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin
First convert 2.5 atm into kPa:
2.5 X 101.3 = 253.25 kPa
Convert 125 Celsius into Kelvin:
125 + 273 = 398 K
Convert Gallons to Litres:
1.25 X 3.79 = 4.74 L
Plug your values into the equation to solve for n:
(253.25)(4.74) = n(8.314)(398)
n = (253.25)(4.74)/(8.314)(398)
n = 0.362 moles
Now use M = m/n to solve for the mass of O2
M = Molar Mass
M = mass
n= moles
32 = m/(0.362)
m = (32)(0.362)
m = 11.58g
Stoichiometry is not really hard if you are good with your numbers. All you have to do is pay close attention and follow you teacher because if you don't you will end up like me not knowing what you are doing. Listen and pay real close attention and I can guarantee you it will be just fine!
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =
Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =
Thus vant hoff factor for sodium chloride in X is 1.9