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MAXImum [283]
2 years ago
14

Movement of electrons about a central nucleus is a concept by whom?

Chemistry
2 answers:
Burka [1]2 years ago
7 0

Answer:

Bohr.

Explanation:

In 1913, N.Bohr proposed an atomic model for atom. He proposed that in an atom there is a central nucleus and electrons move around this center in a fixed path known as shell.

The shells or path have a fixed size and energy. The electrons do not lose energy during this movement.

There is possibility of energy change (emit or absorb) in an electron if they move from one shell to another.

Alenkasestr [34]2 years ago
5 0
The answer is bohr hope this helps :)
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THESE QUESTIONS ARE ELECTRICIAN REALATED
Vladimir [108]

<span>1.      </span>A.

It is a technique typically engaged in electrical power and electronic devices, wherein the devices are run at less than their rated maximum power degeneracy, taking into account the case or body temperature, the ambient temperature and the type of cooling mechanism used.

<span>2.      </span>B.

In order to answer the question, you need to know the area of a circle.<span>
Area = pi * r^2
Area = 3.14 * (100/2) ^ 2
Area = 3.14 * 50^2
Area = 7850 mils^2 = 0.00785 in^2</span>

<span>3.      </span>A.

V / 15 A = 0.213 Ohm <span>
0.213 Ohm / 125 ft * 1000 ft = 1.71 Ohm / 1000 ft 
Closest gauge is #12 AWG (solid wire) 
#12 is 1.588 Ohm / ft, which is a 2.9775 V drop @ 15A</span>

<span>4.      </span>A.

RHW cale insulation can be used up to 167 degrees F.

<span>5.      </span>D.

This is a type of electrical connector used to fasten two or more low-voltage (or extra-low-voltage) electrical conductors.

<span>6.      </span>C. the ease with which a material allows electricity to move is called CONDUCTIVITY

<span>7.      </span>D.

Stranded conductors are not acceptable in the pressure terminals of a duplex receptacle. Aluminum conductors of suitable gauge for a 15 ampere duplex outlet will not fit in the pressure terminal holes. 

<span>8.      </span>D.

A 3/0 AWG THHN insulated Copper is rated at 225 amps and it falls under the 90 degree C table.

<span>9.      </span>D.

A #12 copper conductor with an insulation factor of 90 degrees C is rated at 20 amps. A #12 aluminum conductor with an insulation rating of 90 degrees C is rated at 15 amps. These conductors ratings only applies to three conductors in a rated at 15 amps. These conductors’ ratings only applies to three conductors in a raceway. From 7 to 24 conductors in a raceway, both aluminum and copper conductor's ratings have to be reduced by .70, so 15 amps x .7 = 10.5 amps and 20 amps x .7 = 14 amps respectively.

<span>10.  </span>A.

<span>NEC 310.15(B)(1) and (2), table 310.15(B)(2)(a), and table 310.15(B)(17).
Convert 158°F to C result is 70°C. 50 amps x .33 correction factor = 16.5 amps


</span>

<span>11.  </span>A.

<span>12.  </span>B.

Ideally, nm or armored cable should be installed in a wall stud at least 1 1/4 inches from the front edge of the stud.

<span>13.  </span>C.

Always use flux for electrical connections and never use acid flux on electrical connections

<span>14.  </span>D.

In wire gauge, the lower the number, the larger the diameter. So a 6 gauge is larger than 12 gauge, but smaller than 4 gauge.

<span>15.  </span>A.

<span>16.  </span>D.

<span>17.  </span>C.

Table 310-16 NEC

<span>18.  </span>A <span>
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<span>19.  </span>D.

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<span>The larger the diameter, the larger the ampacity, or the ability to carry current.</span>

4 0
2 years ago
Read 2 more answers
You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu
ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

  • 10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

  • 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

  • 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

  • 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
4 0
1 year ago
A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc
mr Goodwill [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

8 0
2 years ago
10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv
Murljashka [212]

Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate,  KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

Hence, option B is correct

3 0
2 years ago
Read 2 more answers
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seraphim [82]
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g. 

The concentration of the solution in water is,
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Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.
4 0
2 years ago
Read 2 more answers
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