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frosja888 [35]
4 years ago
11

Which of the following elements has the lowest first ionization energy?

Chemistry
2 answers:
Tanya [424]4 years ago
7 0

Answer:

(Sc)

Explanation:

Vsevolod [243]4 years ago
5 0

Answer:

Scandium (Sc)

Explanation:

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Please help!!
viva [34]

Answer:

Answer is 2nd

Explanation:

some energy is transformed into mass

5 0
3 years ago
What process takes place before an article is published in a scientific journal/
Goshia [24]
Before an article is published in a scientific journal or in any "peer-reviewed" journal the article is reviewed thoroughly by scholars from the journal as well as peers or scholars of the articles author from the same field. This process occurs to provide credibility to the ideas being published and so that readers and other scholars can rely on the validity of the material being published. 
7 0
3 years ago
You notice something is growing in a 100ml pot of liquid soap. You take 1ml of this liquid soap and perform a serial dilution an
andrew-mc [135]

Answer:

3'700,000 cfu

Explanation:

One way to count the amount of bacteria in a medium is by doing a dilution of the sample and count how many colonies growth. Each colony is a cfu (Colony forming units).

In the problem, you count 37 colonies. The dilution was 1:100,000. That means the bacteria present in the soap is:

37 colonies × (100,000 / 1) = <em>3'700,000 cfu</em>

<em></em>

I hope it helps!

6 0
3 years ago
Which is an example of making a quantitative observation?
Zolol [24]
Answer:
measuring the mass of metal used in a reaction
3 0
3 years ago
Read 2 more answers
What is the frequency of light with an energy of 124 kJ/mol?
Charra [1.4K]

Answer: = 3.11 x 10^14 s^-1

Explanation:

Use the formula E = hv

This formula uses the assumption that the unit for energy is in Joules/photon.

124 kJ = 124000J

To get 124000J/mol into a unit of J/photons, we need to divide by the number of photons in a mole, which is 6.022 x 10^23.

And thus, we need

124000/6.022 x 10^23 = 2.06 x 10^-19J/photon

We can plug it in to E = hv by

2.06 x 10^-19J = (6.63 x 10^-34 J s)(v) Isolate v by

v = (2.06 x 10^-19J)/(6.63 x 10^-34 J s)

= 3.11 x 10^14 s^-1

6 0
3 years ago
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