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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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Slav-nsk [51]

Answer: The correct answer is (D).

Explanation:

Space photography help astronomers:

  • To visualize and observe the position and appearance of celestial objects like: path of orbiting planets ,dim stars which invisible to naked eyes ,far away galaxies etc.
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5 0
3 years ago
Read 2 more answers
2Al + 3CuSO4–>3Cu+ Al2(SO4)3
sergiy2304 [10]

Answer:

7.82 g of Cu

Explanation:

2 moles of Al react to 3 moles of copper sulfate in order to produce 3 moles of copper and 1 mol of aluminum sulfate.

Firstly we determine the moles of reactant.

As copper sulfate is in excess, Al is the limiting.

2.75 g . 1mol /26.98g = 0.102 moles

Ratio is 2:3. 2 moles of Al, can produce 3 moles of Cu

So the 0.102 moles of Al will produce(0.102 . 3) /2 = 0.153 moles.

We convert moles to mass: 0.153 mol . 63.5g /mol = 9.71 g

That's the theoretical yield (100 % yield reaction)

We know that: (yield produced / theoretical yield) . 100 = percent yield

We replace:

(Yield produced / 9.71g) . 100 = 80.5  %

(Yield produced / 9.71g) = 0.805

Yield produced = 0.805 .  9.71g = 7.82 g

4 0
3 years ago
Calculate the number of moles in each of the following compounds. 180 g of Zn
iVinArrow [24]
You can do this by dividing by the molar mass of Zn:

4 0
3 years ago
I have to show my work after I find the atomic weight of each element please help !!
wariber [46]

Explanation:

For all of the questions, do the following calculation

\frac{isotope \:  \times abundance}{100}  +  \frac{isotope \:  \times abundance}{100} ...

As an example, I'll do question 1

\frac{204 \times 1.480}{100}  +  \frac{206 \times 23.6}{100}  +  \frac{207 \times 22.6}{100}  +  \frac{208 \times 52.3}{100}  = 207.20

5 0
3 years ago
A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane
zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

          \frac{1.5 L}{24 hr}

            = 0.0625 L/hr

As, molar mass of He is 4 g/mol  and molar mass of O_{2} is 32 g/ mol.

Now,

   \frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}

                               = 2.83

or, rate of O_{2} = \frac{\text{Rate of He}}{2.83}

                       = \frac{0.0625 L/hr}{2.83}

          Rate of O_{2} = 0.022 L/hr.

This means that 0.022 L of O_{2} gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of O_{2} gas is calculated as follows.

         \frac{1.5 L}{0.022 L/hr}

                = 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

3 0
3 years ago
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