Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
Answer:
its 2p
Explanation:
In a neutral carbon atom, the 1s sublevel has one orbital with two electrons with opposite spins, represented by the arrows pointing in opposite directions. The 2s sublevel also has one orbital with two electrons, also with opposite spins. The 2p sublevel has three orbitals.
(i don"t understand this i just looked it up ._.
Answer: A
Explanation: i took the test
Answer:
The description of the given question is summarized in the explanation section below.
Explanation:
A maximum of around 82 ppm would be found throughout the beginning material spectrum around CDCl3. This same resulting spectrum maximum of 82 ppm has been lost, as well as the section has a historic high of 215 ppm.
- The OH summit at 82 ppm was obvious as well as a significant maximum of around 215 ppm carboxylic acid was observed.
- That justifies the occurrence of oxidizing, C13 NMR was indeed verified as a reactive organic compound oxidized in carbonyl Ozone.
