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Varvara68 [4.7K]
3 years ago
6

When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory,

Chemistry
1 answer:
rewona [7]3 years ago
3 0

Hey there!

No of hybrid orbitals , H = ( V +S - C + A ) / 2

Where H = no . of hybrid orbitals

V = Valence of the central atom = 5

S = No . of single valency atoms = 4

C = No . of cations = 1

A = No . of anions = 0

For PCl4 +

Plug the values we get H = ( 5+4-1+0) / 2

H =  4 ---> sp3 hybridization

sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations

Answer C

Hope that helps!

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how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
kobusy [5.1K]
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....


Do the calculation
6 0
3 years ago
What percentage of a radioactive species would be found as daughter material after seven half–lives?
irinina [24]
0.78  is what the verified answer says 
6 0
3 years ago
What is the molarity of a solution made by dissolving 1.25 mol of HCl in enough
Ronch [10]

Answer:

Explanation:

C

4 0
3 years ago
Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere
Colt1911 [192]
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M


8 0
3 years ago
EXPERIMENT: THE CABBAGE INDICATOR
Nikitich [7]

Answer:

EXPERIMENT: THE CABBAGE INDICATOR

Here are your goals for this lesson:

Use indicator for each substance to determine if substance is an acid or a base

Summarize results and the experimental method

You can use the natural indicator in red cabbage juice to determine which household liquids are acids or bases. If no change in the color of the cabbage juice takes place, the liquid is neither an acid nor base; it is neutral.

Acids and bases are opposites. When an acids and base are mixed, they neutralize each other. You can use an indicator to see the neutralized reaction. The cabbage juice will change color when the liquids are neutralized.

Online Lab

This video will demonstrate how an indicator, in this case, cabbage indicator, can be used to determine whether a solution is acidic or basic. As you watch the video, remember to record your data and observations to use to present your findings.

Compile a summary of your findings from this experiment. Include your hypothesis, observations, data, and conclusions. Be sure to answer the questions below as well as explaining the method and results.

r.

Explanation:

8 0
2 years ago
Read 2 more answers
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