Why is natural product and secondary metabolism so widely variable in the plant kingdom?

Chemistry

1 answer:

Anna35 [415]2 years ago

8 0

Answer: here u go i hope this hlpe

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Hey! I was wondering how to get the solution to this question!

zzz [600]

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

3 0

3 years ago

There are 31.1 grams (g) in 1 troy ounce (ozt). How many ozt of gold are extracted from 1 short ton of average Nevada ore

devlian [24]

This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

$\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\$