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Likurg_2 [28]
3 years ago
14

What transition energy corresponds to an absorption line at 460 nm?

Chemistry
2 answers:
kykrilka [37]3 years ago
8 0
The answer is b I think I hope it right
galben [10]3 years ago
8 0

Answer:

B. 4.32 x 10-19 J is correct via a p e x

Explanation:

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melamori03 [73]
It changes from a liquid from to a solid.

I hope this helps!
3 0
2 years ago
South carolinians feared that the hire terrace of 1828 would cause
ladessa [460]
They called it the Tariff​ of Abomination aka the Tariff of Evil because it really hurt the south. Any tariffs hurt the South with their agricultural business so I guess they believed it would lead to losing their jobs, etc..
5 0
3 years ago
In the essay box, compose a scientific report which includes your observations, data, and conclusions. The following questions s
timama [110]

Answer:

1. It depends what type of method you are using. if it is Height x Width x Length then it will not work for an irregular shape because it has extra pieces that would not be included.

2. The second method would work for both regular and irregular shapes because you would have to know or find out the volume of the regular shape to get the volume for the irregular shape.

3. It also depends on what you are doing, if you are doing a regular shape then use the first method, if it's an irregular shape then use the second method, if you do the maths correctly both should give you an accurate answer for what you want to achieve.

4. No, because the sugar would dissolve.

5. No, on this case the displacement method would not work because of the weight difference

Explanation:

All the answers for you!

4 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
What are the lanthanide & actinide series?
ch4aika [34]

Answer:

 

The lanthanides and actinides together are sometimes called the inner transition elements.

Explanation:

They are called this because they come up in the periodic table after actinium

Hope this helps :)

8 0
2 years ago
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