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Kaylis [27]
3 years ago
15

What would be another property that we Earthlings could use to classify stars?

Chemistry
1 answer:
Blizzard [7]3 years ago
8 0

Answer: At the point when space experts take a gander at an article's range, they can decide its arrangement dependent on these frequencies. The most well-known technique stargazers use to decide the sythesis of stars, planets, and different articles is spectroscopy. This spread-out light is known as a range.

Explanation:

You might be interested in
What are specatator ions
Margaret [11]

Answer:

A spectator ion is an ion that does not participate in a chemical reaction and is found in a solution before and after a reaction.

Hope this helped :)

5 0
2 years ago
What is the ph of a solution that is 0.100 m hio3and 0.100 m naio3?
Nookie1986 [14]
What are the answer choices?
8 0
3 years ago
Which pair of elements are in the same period?Hg & PbLi & NaB & ClO & As
Alexeev081 [22]

Answer:

Hg Pb are in one, Li and B and O are in one, and NaCl are in one, As is alone

Explanation:

Periods are horizontal

Hope this helps!

4 0
2 years ago
!!HELP PLS!! A compound sample contains 0.783g of C, 0.196 g of H, 0.521 g 0 and the molecular formula molar mass is 184.27g/mol
Delvig [45]

Answer:

First

divide each element by its Molecular Mass to get their respective moles

Then Divide through by the lowest of the moles

You'll have the ratio of Carbon Hydrogen and Oxygen to be

C2H3O

Given Molecular Mass=184.27

C2H3On=184.27

n(12x2 + 1x3 + 16) =184.27

Evaluating this... You'll have n=4.3

Pls check if you assigned the correct value to each element

4 0
2 years ago
Calculate the standard emf for the following reaction:
krek1111 [17]
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l) 
</span><span>E = +1.47 
</span>
<span>Br(l) + 2e- = 2Br- 
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
4 0
3 years ago
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