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antoniya [11.8K]
3 years ago
9

The impacts of global warming are being felt and will be felt increasingly in your lifetime. What impacts are likely to be seen

in the next few decades
Chemistry
1 answer:
IceJOKER [234]3 years ago
8 0

Answer:

Hotter climate, more greenhouse gases, and less water

I hope this helps :)

Explanation:

You might be interested in
A container of hydrogen gas has the same temperature as a container of oxygen gas. The gas atoms having the greater average kine
user100 [1]

Answer:

hope this helps

Explanation:

The atoms of hydrogen have smaller mass than oxygen. Thus their speeds have to higher in order to produce the same average kinetic energies.

4 0
2 years ago
After he conducted cathode ray tube experiments proving the existence of negatively charged particles we now call electrons, Tho
Lina20 [59]

Answer:

Answer is explained below;

Explanation:

In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.

In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.

By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.

He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.

So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.

Later, the Rutherford model was replaced by the Bohr atomic model.

6 0
3 years ago
Name the 3 seperate metals group
umka21 [38]

Akali Metals

Akali-Earth Metals

and Other Metals

8 0
2 years ago
2
Alina [70]

Answer:

N - 1s²2s²2p³

Explanation:

Nitrogen is located in the p-block of the periodic table (groups 13-18) and is on the 2nd period.

The 2nd period tells us the principal energy level (a quantum number) is n = 2. Therefore, it must have already filled up the 1s sublevel.

The groups 13-18 on period 2 tells us that the 2s sublevel is also filled.

Nitrogen is located in Group 15. That means that there are 3 electrons that have filled the 2p sublevel, out of a possible 6.

Therefore, our electron configuration is 1s²2s²2p³

2p³ (Shorthand Config)

[He] 2s²2p³ (Noble Gas Config)

7 0
3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
2 years ago
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