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2 years ago

An electron is on a -2.5 eV energy level. the electron is stuck by a 2.5 eV photon. What will most likely happen?

Chemistry

1 answer:

Luba_88 [7]2 years ago

3 0

The answer would be 0

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For scientific notation, when i move the decimal to the left the the exponent get bigger or smaller?

xeze [42]

I'm not sure what you mean. Besides, I feel like you're talking math.
But anyways, if you have 120 let's say
The scientific notation is 1.20 × 10^2
if you have 125000
the scientific notation is 1.25 × 10^ 5
The number of times you go left the decimal, I guess exponent increases
So yea

3 0

2 years ago

In the laboratory you dissolve 12.3 g of chromium(ii) sulfate in a volumetric flask and add water to a total volume of 375. ml.

valentinak56 [21]

Hope it cleared your doubt.

4 0

3 years ago

Which of the following are advantages of the SI system?

lions [1.4K]

The correct answer is A, B and C

3 0

3 years ago

Read 2 more answers

What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientifi

Tomtit [17]

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

$V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL$

3 0

3 years ago

a researcher obtains a sample of 0.070 M nitrate solition. A 20.0 mL aliquote of the nitrate solution is added to 10.0 mL of amm

katovenus [111]

Answer:

Concentration of nitrate in the new solution = 0.007 M

Explanation:

Given:

Concentration nitrate solution = 0.070 m

Volume of aliquote of the nitrate solution is add = 10.0 ml

Total volume = 100 ml

Find:

Concentration of nitrate in the new solution

Computation:

Number of M. mole = 0.070 m x 10.0 ml

Number of M. mole = 0.7 m-moles

Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml

Concentration of nitrate in the new solution = 0.007 M

6 0

2 years ago