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tino4ka555 [31]

2 years ago

15

when an electron moves from a higher orbit to a lower one does it always follow the same path explained

2 answers:

Ilia_Sergeevich [38]2 years ago

8 0

Yes, electron follows the same path when it absorb and loses energy.

Yes, when an electron moves from a higher orbit to a lower orbit it always follow the same path as it moves from a lower orbit to a higher orbit. When electron absorb energy it has the power to move from lower orbit to higher orbit or energy level.

While on the other hand, when an electron loses that energy, it comes back to its original position from which it moves earlier when it absorb energy so we conclude that electron follows the same path when it absorb and loses energy.

Learn more: brainly.com/question/24962163

morpeh [17]2 years ago

7 0

Answer:

Every time an electron changes its orbit, from a higher energy one to a lower energy one, it gives off a photon of light whose energy is the difference in the energy between the two states. This is the big success of the Bohr atom.

Explanation:

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

25. 00 ml of a buffer solution contains 0. 500 m hclo and 0. 380 m naclo. if 50. 00 ml of water is added to the buffer, what are

geniusboy [140]

The new concentrations of $HClO$ and $NaClO$ are 0.25M and 19M

Calculation of number of moles of each component,

Molarity of $HClO$ = number of moles/volume in lit = 0. 500 M

Number of moles = molarity of $HClO$ × volume in lit = 0. 500 M× 0.025 L

Number of moles of $HClO$ = 0.0125 mole

Molarity of $NaClO$ = number of moles/volume in lit = 0. 38 M

Number of moles = molarity of $NaClO$ × volume in lit = 0. 38 M× 0.025 L

Number of moles of $NaClO$ = 0.95 mole

Calculation of new concentration at volume 50 ml ( 0.05L)

Molarity of $HClO$ = number of moles/volume in lit = 0.0125 mole/0.05L

Molarity of $HClO$ = 0.25M

Molarity of $NaClO$ = number of moles/volume in lit = 0.95mole/0.05L

Molarity of $NaClO$ = 19 M

learn about Molarity

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1 year ago

How many atoms are present in 179.0 g of iridium?

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160 atoms are present

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Lipids cannot dissolve in water.

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3 years ago

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I know there are 4.5*10^-9 seconds in 4.5 nanoseconds, but when I enter it into my TI-30xIIs, I get 4.5*10^-10. I enter it as 4.

o-na [289]

Answer:

You are not using properly the function exponential in your calculator

Explanation:

When a number is too big or too small we use scientific notation. This is a number between 1 a 10 multiplied by a power of 10.

When you are writing 4.5*10^-9 you are actually writing 0.0000000045 in scientific notation.

When you enter this in the calculator you have to use the function EXP after the first two numbers.

Steps: 1) Enter 4.5

2) Enter EXP

3) Enter minus (-)

4) Enter 9

8 0

3 years ago