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sergejj [24]
3 years ago
11

In the 1800s, a popular belief known as vitalism stated that life processes could not be explained by the laws of physics and ch

emistry, and were instead dictated by an independent life force. Which discovery most likely caused scientists to revise this hypothesis regarding the origin of life on Earth
Chemistry
1 answer:
soldi70 [24.7K]3 years ago
6 0

The question is incomplete, the complete question is;

In the 1800s, a popular belief known as vitalism stated that life processes could not be explained by the laws of physics and chemistry,and were instead dictated by an independent life force. Which discovery most likely caused scientists to revise this hypothesis regarding the origin of life on Earth?

a. that inorganic compounds existed within live organisms

b. that organic compounds could be synthesized in a laboratory

c. that RNA could serve as a template to synthesize DNA

d. that self-replicating molecules existed inside cells

Answer:

b. that organic compounds could be synthesized in a laboratory

Explanation:

Vitalism is the belief that "living organisms are fundamentally different from non-living entities because they contain some non-physical element or are governed by different principles than are inanimate things"(wikipedia).

This theory held that the molecules involved in life processes could not be synthesized in the laboratory.

All these were upturned after Fredrich Whöler's synthesis of urea in 1828. He was able to show that molecules involved in life process can also be synthesized in the laboratory. This gave rise to modern synthetic organic chemistry.

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Calculate the percent ionization of nitrous acid in a solution that is 0.222 M in nitrous acid (HNO3) and 0.278 M in potassium n
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Explanation:

Concentration of KNO_{2} is 0.278 M and it is completely ionized into K^{+} and NO^{-}_{2}.

This means that [KNO_{2}] = [NO_{2}] = 0.278 M

It is given that concentration of HNO_{2} is 0.222 M.

As HNO_{2} is a weak acid. Therefore, its dissociation will be as follows.

              HNO_{2}(aq) \rightarrow H^{+}(aq) + NO^{-}_{2}(aq)

Initially :    0.222 M            0        0.278 M

Change :    - x                    +x           +x

Equilibrium : 0.222 M - x    x         0.278 M + x

Therefore, dissociation constant for this reaction will be as follows.

        K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}

Hence, putting the given values into the above formula as follows.

        K_{a} = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}

      4.50 \times 10^{-4} = \frac{x \times (0.278 + x)}{(0.222 - x)}

                          x = 0.00036

As, [H^{+}] is 0.00036 M. Therefore, percentage ionization of HNO_{2} will be calculated as follows.

              % ionization of HNO_{2} = \frac{[H^{+}]}{[HNO_{2}]} \times 100

                               = \frac{0.00036 M}{0.222 M} \times 100

                               = 0.16 %

Thus, we can conclude that % ionization of HNO_{2} is 0.16%.

5 0
4 years ago
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