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sukhopar [10]
3 years ago
5

How many valance electrons are in group 16

Chemistry
2 answers:
labwork [276]3 years ago
7 0

Answer:

There are 6 valence electrons in group 16..

Mark me as brainliest if u found this answer helpful

Natalija [7]3 years ago
3 0

Answer:

6

Explanation:

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Please hep its my last question
atroni [7]

Answer:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

Explanation:Noble gases:

 are highly reactive.

 react only with other gases.

 do not appear in the periodic table.

 are not very reactive with other elements.

7 0
2 years ago
if the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
Makovka662 [10]
The answer to this question is:

<span>If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
</span><span>"Claim significant support for the hypothesis"

Hoped This Helped, </span><span> Itsalishamariee
Your Welcome :)</span>
6 0
2 years ago
Read 2 more answers
What is the SI base unit for length?
sp2606 [1]

The SI base unit for length is meter.

In order to make smaller measurements, you can use the centi-, milli-, micro-, etc. prefixes.

When you want to reference larger measurements, you can use the kilo-, mega-, giga- and prefixes such as those.

5 0
3 years ago
Calculate the value of AE for a system that loses 50 J of heat and has 150 J of work Performed on it by the surroundings.
gavmur [86]

Answer:

100 j

Explanation:

bascially u j subtract 50 from 150 because the system lost 50 j but theres 150 in its surroundings

4 0
2 years ago
What is the hybridization and shape for an XeF6 2+ molecule?
Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)      

8 0
3 years ago
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