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Naddik [55]
3 years ago
6

write the ground-state electron configurations of the seventh and eighth elements in the first transition series in Universe fif

th transition metal: sixth transition metal:
Chemistry
1 answer:
Tju [1.3M]3 years ago
5 0

Answer: Check explanation.

Explanation:

Transition metals are metallic elements that can be found in the Groups IVB–VIII, IB, and IIB on the periodic chart.

The question require us to write down the GROUND state electronic configuration of the first fifth transition metal, sixth transition metal, seventh transition metal element and the eighth transition metal.

NOTE: we are Starting from Argon, which has 18 electrons.

The fifth transition metal is Manganese,Mn. Manganese has 25 electrons, that is, 25- 18= 7. Therefore, it needs seven electrons to complete the configuration.

Hence, The ground state electronic configuration = [Ar)] 3d5. 4s2.

The first sixth Transition metal is iron,Fe. Iron has 26 electrons, that is, 26 - 18 = 8. Therefore, it need eight Electrons to complete the ground state electronic configuration.

Hence, the ground state electronic configuration of Fe= [Ar] 3d6. 4s2.

The first seventh transition metal is Cobalt, Co. It has 27 Electrons, therefore, 27- 18 = 9. Therefore, it needs 9 Electrons to complete its ground state electronic configuration.

Ground state electronic configuration of Co= [Ar] 3d7. 4s2.

The first eight Transition metal is Nickel. It has 28 electrons. Therefore, 28-18= 10. So, it needs 10 Electrons to complete its ground state electronic configuration.

Hence, the Ground state electronic configuration of Ni= [Ar] 3d8. 4s2.

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A plate moves 200 m in 10,000 years. What is its rate in cm/year?
Varvara68 [4.7K]

Answer:

The answer is

<h2>2 cm/year</h2>

Explanation:

To find the rate in cm/year we must first convert 200 m into cm

1 m = 100 cm

if 1 m = 100 cm

Then 200 m = 200 × 100 = 20 ,000 cm

So the rate is

<h2>\frac{20000}{10000}</h2>

<u>Reduce the fraction with 10,000</u>

We have the final answer as

<h3>2 cm/year</h3>

Hope this helps you

3 0
3 years ago
Salt water contains n sodium ions (NA ) per cubic meter and n chloride ions (Cl-) per cubic meter. A battery is connected to met
IRINA_888 [86]

Answer: hello your question is incomplete below is the complete question

Salt water contains n sodium ions (Na+) per cubic meter and n chloride ions (Cl−) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt water. The cross sectional area of the pipe is A. The magnitude of the drift velocity of the sodium ions is VNa​ and the magnitude of the drift velocity of the chloride ions is VCl​.

What is the magnitude of the ammeter reading ?

answer :

| I | = neAVₙₐ  + neAV (Cl-)

Explanation:

Given that there are N sodium ions

<u>Determine the Magnitude of the ammeter reading </u>

| I | = current due to sodium ions + current due to (Cl-) ions

     = neAVₙₐ  + neAV (Cl-)

3 0
3 years ago
What is the basic SI unit of time
aniked [119]

Answer:

The base unit for time is the second the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance!

Explanation:

I hope i helped =)

5 0
3 years ago
In a molecule of methane, carbon (having four valence electrons) binds to four hydrogen atoms. How many lone pairs of electrons
antiseptic1488 [7]
None. To bind with an hydrogem atom, the carbon needs to ''share'' one of it's valence electron's with the hydrogen's valence electron. Therefore, forming 4 bounds means the carbon atom will share it's 4 valence electrons and there will be no lone electrons.
7 0
3 years ago
In a coffe cup calorimeter, 50.0mL of 0.100M of AgNO3 and 50mL of 0.100M HCl are mixed to yield the following reaction:
Jet001 [13]

Answer:

The enthalpy change of the reaction is -66.88 kJ/mol.

Explanation:

Mass of the solution = m = 100 g

Heat capacity of the solution = c = 4.18 J/g°C

Initial temperature of the solutions before mixing = T_1=22.60^oC

Final temperature of the solution after mixing = T_2=23.40^oC

Heat gained by the solution due to heat released by reaction between HCl and silver nitrate = Q

Q=m\times c\times (T_2-T_1)

Q=100 g\times 4.18 J/g^oC\times (23.40^oC-22.60^oC)=334.4 J

Heat released due to reaction = Q' =-Q = -334.4 J

Moles of silver nitrate = n

Molarity of silver nitrate solution = 0.100 M

Volume of the silver nitrate solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

Moles =Molarity\times Volume (L)

n=0.100 M\times 0.050 L=0.005 mol

Enthalpy change of the reaction = \Delta H

=\Delta H=\frac{-334.4 J}{0.005 mol}=-66,880 J/mol=-66.88 kJ/mol

1 J = 0.001 kJ

The enthalpy change of the reaction is -66.88 kJ/mol.

7 0
3 years ago
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