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AlexFokin [52]
3 years ago
11

How many orbitals surround the nucleus in a neutral atom of sulfur (S)?

Chemistry
1 answer:
Leto [7]3 years ago
3 0
<h2>C</h2>

Explanation:

The atomic number of S is 16

So,number of electrons in S is 16

The electronic configuration of S is 2,8,6

The orbital electronic configuration of S is ^{1}s_{2}^\text{ }^{2}s_{2}^\text{ }^{2}p_{x2}^\text{ }^{2}p_{y2}^\text{ }^{2}p_{z2}^\text{ }^{3}s_{2}^\text{ }^{3}p_{x2}^\text{ }^{3}p_{y1}^\text{ }^{3}p_{z1}^\text{ }

So,the number of orbitals involved is 9.

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Alex777 [14]
I’m sorry I didn’t understand can you please add more details please thank you
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4 years ago
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A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution after the addition of 100.0 ml
Naddika [18.5K]
(I think you have a mistake in your question as the addition is 30mL, not 100mL)
when PH = - ㏒[H+]
and here we have HClO4 is the strong acid
So PH = - ㏒[HClO4]

moles of HClO4 = 0.1 L *0.18 m = 0.018 M
moles of LiOH = 0.03 L * 0.27 m = 0.0081 M
when the total volume = 0.1L + 0.03L =  0.13 L
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5 0
4 years ago
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A car at rest ends accelerates for 12 seconds. After this time the car is going 36 m/s. What was its acceleration?
Zigmanuir [339]

Acceleration is defined as velocity per unit time.

Acceleration=\frac{Velocity}{time}

a=\frac{dv}{dt}

Here, a=acceleartion,

v=velocity=36 m/s

t=time=12 s

a=\frac{dv}{dt}

a=\frac{36}{12}

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A car at rest ends accelerates for 12 seconds. After this time the car is going 36 m/s. So acceleration that is a=3 ms⁻².


4 0
3 years ago
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How many moles of air are necessary for the combustion of 5.00 molmol of isooctane, assuming that air is 21.0% O2O2 by volume?
Annette [7]

Answer:

67.5 moles of O2

Explanation:

2  C 8 H 18  +  27 O 2  →  16 C O 2  +  18 H 2 O

According to the balanced chemical equation, in normal air % moles of iso-octane will require 67.5 moles of oxygen to combust completely...

8 0
4 years ago
Convert 145 meters to kilometers.
Sphinxa [80]

Divide by 1000

145÷1000=0.145

Hope this will be helpful.

4 0
1 year ago
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