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Goryan [66]
3 years ago
10

Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon

stration, through the combined work of Joseph Priestley, Jan Ingenhousz, and Jean Senebier, that photosynthesis is essentially the reverse of respiration was a milestone in the development of biochemical thought.
Chemistry
1 answer:
balu736 [363]3 years ago
3 0

Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

Thus, at the end of the century the participation of the atmosphere in plant dynamics was already seated, although the how and why of this participation were still unknown and no theory had been formulated to explain the nutritional process as a whole.

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For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
Imagine mixing 1 tablespoon of Epsom salt with 2 cups of ammonia. How much precipitate would be produced? Describe the amount of
iVinArrow [24]

Answer:

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjkwv-cqrjnAhVCheAKHWaFBBgQFjAAegQICBAB&url=https%3A%2F%2Fwww.acs.org%2Fcontent%2Fdam%2Facsorg%2Feducation%2Fresources%2Fk-8%2Finquiryinaction%2Fstudent-activity-sheets%2Fgrade-5%2Fchapter-3%2Flesson-3.3-forming-a-precipitate.pdf&usg=AOvVaw1fT7fpXG9PNWroM87puvgQ

Explanation:

that has the answers copy and paste it in your google

6 0
3 years ago
Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
3 years ago
The _____ method of pest control combines chemical and non-chemical methods and emphasizes the use of natural toxins.
labwork [276]
The answer to the question stated above is:
Integrated Pest Management is a method of agricultural pest control that combines chemical and non <span>chemical methods and emphasizes the use of natural toxins and effective pesticide.

>>N</span><span>atural toxins in plants can also be isolated and applied to crops like synthetic pesticides.</span>
4 0
3 years ago
Read 2 more answers
If is found that 24.68 mL of .1165 M NaOH is needed to titrate .2931 g of an unknown acid to the phenolphthalein end point. Calc
Lemur [1.5K]

Answer: The equivalent mass of the acid is 83.16 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}  

Molarity of NaOH solution = 0.1165 M

Volume of NaOH solution = 24.68 mL = 0.02468 L

Putting values in equation 1, we get:

\text{Moles of} NaOH={0.1165M}\times{0.02468L}=2.875\times 10^{-3}moles=2.875\times 10^{-3}geq    

( as acidity of NaOH is 1)

For end point:  gram equivalents of acid =  gram equivalents of base = 2.875\times 10^{-3}

Mass of acid=gram equivalents\times {\text {Equivalent mass}}

0.2391=2.875\times 10^{-3}\times {\text {Equivalent mass}}

{\text {Equivalent mass}}=83.16g

Thus equivalent mass of the acid is 83.16 grams

5 0
3 years ago
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