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oksano4ka [1.4K]
3 years ago
12

Which answer lists the fundamental forces in order from strongest to weakest

Chemistry
2 answers:
laiz [17]3 years ago
8 0
Actually, gravity is the weakest of the four fundamental forces. Ordered from strongest to weakest, the forces are 1) the strong nuclear force, 2) the electromagnetic force, 3) the weak nuclear force, and 4) gravity.May 22, 2013
Zanzabum3 years ago
6 0

Answer:

C strong, electromagnetic, gravity.

Explanation:

if youre for connexus i got you.

the other person just copied and pasted from this

https://wtamu.edu/~cbaird/sq/2013/05/22/why-is-gravity-the-strongest-force/

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10. Which of the following best describes the reaction 2VO3– (aq) + Zn (s) + 8H+ (aq) → 2VO2 (aq) + Zn2+ (aq) + 4H2O (l)? ______
Oduvanchick [21]
Correct answer is option E. <span>It is a redox reaction in which Zn is oxidized at the anode, and V is reduced at the cathode.

Reason:
In above reaction, the oxidation state of VO3- is +5, while that of VO2 is +4. Thus there is reduction of V from +5 to +4
In case of Zn, oxidation state of Zn is increased from 0 to +2, Thus process is referred as oxidation. </span>
3 0
2 years ago
19. What is the density of a 16.39
slavikrds [6]
Density= mass/volume

16.39g/18.00mL = 0.9105 g/mL
Make sure to use the correct number of significant figures (4)
7 0
3 years ago
Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon
balu736 [363]

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.

3 0
3 years ago
Household items that are substances
Yuri [45]

Answer:

water,salt are the house hold items

4 0
3 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
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