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AleksandrR [38]
2 years ago
5

What are the two elements that are liquids at freezing?

Chemistry
2 answers:
-Dominant- [34]2 years ago
8 0

Answer:

I belive 2 of them are Bromine and Mercury there are more then just 2 but im not the best at science sorry.

Explanation:

TEA [102]2 years ago
6 0
Answer: Two elements are liquid at standard temperature and pressure: mercury and bromine. All the other elements are solid under standard conditions, which means their freezing point is above 0 degrees Celsius. Hope this helps (:
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When an ionic bond forms between lithium (Li) and fluorine (F), which of the following occurs?
allochka39001 [22]
They don't share and both deal with only 1 electron.

Lithium gives away 1 electron.
F will receive that electron.

The chemical formula is LiF
3 0
3 years ago
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Two masses exerting a force on each other is an example of What????
trasher [3.6K]

Newtons universal law of gravitation

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8 0
3 years ago
Identify each of the following sets of quantum numbers as allowed or not allowed in the hydrogen atom.
Westkost [7]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

1s^1

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.

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5 0
2 years ago
The red balls represent _____ and have a positive charge.
AysviL [449]
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4 0
3 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
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