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Simora [160]
2 years ago
8

In the compound k[co(c2o4)2(h2o)2] (where c2o42– = oxalate) the oxidation number and coordination number of cobalt are, respecti

vely
Chemistry
2 answers:
anygoal [31]2 years ago
8 0

Answer:

Oxidation number: 3+

Coordination number: 4

Explanation:

<em>In the compound K[Co(C₂O₄)₂(H₂O)₂] (where C₂O₄²⁻ = oxalate) the oxidation number and coordination number of Cobalt are, respectively?</em>

<em />

The sum of the oxidation numbers and charges of the species forming the complex is equal to the overall charge, in this case, zero.

K + Co + 2 . C₂O₄²⁻ + 2 . H₂O = 0

+1 + Co + 2 . (2-) + 2 . 0 = 0

Co = 3+

The coordination number is the number of ligands attached to the central atom. 2 oxalates and 2 water molecules are attached to the cobalt atom, so the coordination number is 2 +2 = 4.

ehidna [41]2 years ago
6 0
Let's Assign Symbols to molecules like,

                    C₂O₄  =  X
and 
                    H₂O   =  Y
Then,
                                K [ Co (X)₂ (Y)₂ ]

As, Potassium (K) has a O.N  =  +1

To neutralize, the coordination sphere must have -1 oxidation number.
So,
                                    [ Co (X)₂ (Y)₂ ]  =  -1
As,
           O.N of X  = -2
Then 
       O.N of (X)₂  =  -4

Also,
O.N of H₂O is zero as it is neutral, So,

                                    [Co - 4 + 0 ]  =  -1
Or,
                                    Co  =  -1 + 4

                                    Co  =  +3

Result:
          Oxidation Number of Coordination Sphere is -1 and Oxidation Number of Cu is +3.
You might be interested in
Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi
gladu [14]

Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

Ptotal = PO2 + PN2 + PHe

= 55 + 90 + 50

= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

8 0
2 years ago
A fictional element has two isotopes and an atomic mass of 87.08 amu. if the first isotope is 86 amu and the second isotope has
Lorico [155]
The atomic mass of a certain element is summation of the product of the decimal equivalent of the percentage abundance and the given atomic mass of each of the isotope. If we let x be the percentage abundance of the 86 amu-isotope then, the second one is 1-x such that,
                      x(86) + (1 - x)(90) = 87.08
The value of x from the equation is 0.73. This value is already greater than 0.5. Thus, the isotope with greatest abundance is that which is 86 amu. 


4 0
2 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
Simplify: <br>(100 m)/(26 s)
podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

4 0
2 years ago
Read 2 more answers
How many chloride ions are in 0.486 moles of chloride ions?​
svetlana [45]

Answer:

Since in a chloride ion, we have an additional electron

you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

Amount of ions in 1 mol = 6.022 * 10^23

Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)

Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

Hence, option 1 is correct

6 0
2 years ago
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