Organic chemistry too eazy but why 90% students avoid ?

Chemistry

1 answer:

Kruka [31]2 years ago

7 0

Answer:

because it covers a large area as there are more then 1 lakh compounds of organic chemistry.

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HeLpPp asap I’ll mark you as brainlister

jeka57 [31]

It’s 1 or 3 sorry if this is wrong

8 0

2 years ago

Write the Henderson-Hasselbalch equation for a propanoic acid solution (CH3CH2CO2H, pKa = 4.874) using the symbols HA and A–, an

Luden [163]

The Henderson-Hasselbalch approximation is for conjugate acid-base pairs in a buffered solution. We're going to call HA a weak acid, and A- its conjugate base. The equation is as follows:
pH = pKa + log([base]/[acid]), where the brackets imply concentrations
Plugging in our symbols and the pKa value, the equation becomes:
pH = 4.874 + log([A-]/[HA])

7 0

2 years ago

molecules with a plane of symmetry between the chiral centers are achiral and meso. identify which of the following molecules ar

maks197457 [2]

Molecules with a plane of symmetry between the chiral centers are achiral and meso. From the given molecules (Picture attached) only (A) compound 1 is meso.

When compounds possess a plane of symmetry between the chiral centers they are called achiral or meso compounds. Among the given compounds (A) compound 1 have a plane of symmetry. So we can say compound one is a meso or achiral compound. Compounds two, three, and four have no plane of symmetry, as you can see in the structures attached. So all other compounds (compound 2, compound 3, and compound 4) except compound one are not meso or achiral.

You can also learn about meso compounds from the following question:

brainly.com/question/29022658

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7 0

7 months ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

2 years ago

Sugar crystals aren't very big when they stop growing. how could you make bigger sugar crystals?

hram777 [196]

DONT GIVE THEM WATER OR HEAT

7 0

2 years ago