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Fed [463]
2 years ago
15

The elements least likely to form bonds are found in which group? 15, 16, 17, 18

Chemistry
2 answers:
Alexandra [31]2 years ago
8 0

Answer:

18

Explanation:

Just took the test

Ipatiy [6.2K]2 years ago
3 0

Answer:

D.18

Explanation:

Edgenuity

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What does NO5 stand for?
nekit [7.7K]

1- Nitrogen 5- Oxygen maybe, i can't find it on the periodic table

3 0
3 years ago
Explain preparation of alkanes C--C from industrial source for industrial preparation of alkane​
Dmitrij [34]
Alkanes can be prepared from carboxylic acid via the removal of carbon dioxide. This process is known as decarboxylation. It produces alkane with a carbon atom lesser than that present in the carboxylic acid.
4 0
2 years ago
A gas has a volume of 100 mL and a pressure of 6 atm. What volume will the gas occupy when the pressure is reduced to 1 atm (tot
serious [3.7K]

<u>We are given:</u>

V1 = 100 mL          P1 = 6 atm

V2 = x mL             P2 = 1 atm

<u>Solving for 'x' : </u>

According to the Boyle's law:

P ∝ 1/V           (pressure and volume are inversely proportional)

PV = k            (where k is a constant)

since the constant k will be the same:

P1V1 = P2V2

replacing the variables

6 * 100 = 1 * x

x = 600 mL   OR   0.6L

Therefore, the gas will have a volume of 600mL or 0.6L

8 0
3 years ago
Suppose a mountain range is made up of 18 mountains, each mountain has
8_murik_8 [283]

The answer is 1.98 x 10^10 leaves.

18 mountains x 8800 trees = 158400 total trees

158400 total trees x 125000 leaves = 19800000000 total leaves in entire mountain range

Sci not: 1.98 x 10^10 leaves  

To do scientific notation, put the decimal place right after the first digit in the number and count how many places are behind the decimal. you'll write the number of places after the decimal after the ^ in the 10^.

7 0
3 years ago
A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole
erastovalidia [21]

Explanation:

Formula for black body radiation is as follows.

                 \frac{P}{A} = \sigma \times T^{4} J/m^{2}s

where,       P = power emitted

                 A = surface area of black body

          \sigma = Stephen's constant = 5.6703 \times 10^{-8} watt/m^{2}.K^{-4}

As area is given as 1.0 cm^{2}. Converting it into meters as follows.

           1.00 cm^{2} \times \frac{(10^{-2})^{2} m^{2}}{1 cm^{2}}      (as 1 m = 100 cm)

            = 1 \times 10^{-4} m^{2}

It is given that P = 201 watts. Hence,  

       \frac{201 watts}{1 \times 10^{-4} m^{2}} = 5.6703 \times 10^{-8} watt/m^{2}.K^{-4} \times T^{4}

                T^{4} = 35.45 \times 10^{12}

                       T = (35.45 \times 10^{12})^{1/4}

                          = 8862.5 K

Thus, we can conclude that the temperature of the surface is 8862.5 K.

7 0
3 years ago
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