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katen-ka-za [31]
3 years ago
5

How many orbitals are available in a “d” orbital configuration?

Chemistry
1 answer:
ratelena [41]3 years ago
8 0

Answer:

e.)5

Explanation:

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Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.
Harlamova29_29 [7]
Following reaction is involved in above system
HOCl(aq)  ↔  H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l)  ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K  = [OCl-][H+<span>]/[HOCl]   ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>)   ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K  = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14

</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
6 0
3 years ago
Read 2 more answers
What best describes the collision between ideal gas molecules
Alika [10]
An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces. One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other.

Happy to help
6 0
3 years ago
What is the mole ratio of NO2 to O2
Mazyrski [523]
The mole ratio is 4 NO2 to 3 O2; 4:3
8 0
3 years ago
The molecule NH3 contains all single bonds.<br><br> true <br> false
dsp73
<h3><u>Answer;</u></h3>

True

<h3><u>Explanation</u>;</h3>
  • The molecule NH3 contains all single bonds.
  • NH3 has a three single covalent  bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
  • There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.
8 0
3 years ago
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How many milliliters of a 3.0 M HCL solution are required to make 250.0 millimeters of 1.2 M HCL?
White raven [17]
The problem above can be solved using M1V1=M2V2  where M1 is the concentration of the concentrated, V1 is the volume of the concentrated solution, M2 is the concentration of the Dilute Solution, V2 is the Volume of the dilute solution. Hence,

(3.0 M)(V2)=(250 mL)(1.2M)
V2 (3.0)= 300
V2= 100 mL

Therefore, you need 100 mL of 3.0 M HCl to form a 250 mL of 1.2 M HCl.
7 0
3 years ago
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