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antiseptic1488 [7]
3 years ago
13

What is the correct order for the reactions that produce the following transformation. a. (1) H2/Lindlar (2) CH3CO2OH b. (1) H2/

Lindlar (2) O3, Zn, HCl c. (1) H2/Pd (2) CH3CO2OH d. (1) Na, NH3 (2) CH3CO2OH
Chemistry
1 answer:
Basile [38]3 years ago
4 0

Answer:

Explanation:

Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)

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A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

q=502J

Hence, the joules of heat were re-leased by the lead is 502

5 0
3 years ago
(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n
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Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

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According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

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0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

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