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sp2606 [1]
3 years ago
12

How many unpaired electrons are in the following complex ions?

Chemistry
1 answer:
Oduvanchick [21]3 years ago
8 0

Solution :

A complex ion may be defined as a metal ion that is located at the center with other molecules or ions surrounding the center ion.

In the context, the number of the unpaired electrons for the following complex ions are :

1. $Ru(NH_3)62+$  (low spin)

  $Ru^{2+} = [Kr] \ 4d^6 \ 5s^0$

   Here all the d-electrons are paired. Therefore, there are no unpaired electrons.

2. $Ni(H_2O)62+$  (high spin)

   $Ni^{2+} = [Ar] \ 4s^0 \ 3d^8$

   Therefore, it has two unpaired electrons.

3. $V(en)33+$  (high spin)

    $V=[Ar] \ 3d^2 \ 4s^0$

   It has 2 unpaired electrons

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It is the smallest planet found in the habitable zone.

Explanation:

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What is the speed of a cheetah that runs 30 miles in 0.5 hours?<br> a
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60 mph (miles per hour)

Explanation:

0.5 hours is 1/2 of an hour, so to get the number of miles for a whole hour you multiply the miles ran by 2.

30 times 2 is 60.

6 0
2 years ago
Read 2 more answers
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.
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Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g  for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

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number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

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How does e=mc^2 connect to quarks
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Answer:

good bro

Explanation:

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