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rjkz [21]
3 years ago
8

There are three competing factors at play here: 1. The effective nuclear charge 2. The size of the atom and the force of attract

ion according to Coulomb's law 3. A pair of electrons in an orbital The first ionization energy is the energy required to completely remove the first electron from the atom. The higher in energy an electron is to start with, the less additinal energy will be required to remove it, which translates to a lower ionization energy. By the same token, an electron arrangement which is lower in energy will require more energy to remove an electron.
Chemistry
1 answer:
alexandr402 [8]3 years ago
3 0

Answer:

These three factors are required for ionization potential or ionization energy.

Explanation:

Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.

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Which principle goes against Bohr's theory and why??​
Basile [38]

Answer:

As the Bohr's fixed orbit gives precise information about the radial position and momentum of the orbit, it is against the Heisenberg uncertainty principle. Thus it is inferred that the Heisenberg uncertainty principle goes and the concept Bohr's fixed Orbit are opposite to each other.

Explanation:

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3 years ago
Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro
melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0
3 years ago
Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt
oksian1 [2.3K]

Answer:

The standard cell potential is 2.00 V

Explanation:

<u>Step 1:</u> Data given

Cu is cathode because of higher EP

Al3++3e−→Al       E∘=−1.66 V     anode

Cu2++2e−→Cu    E∘=0.340 V    cathode

<u>Step 2:</u> Balance both equations

2*(Al → Al3+-3e−)       E∘=1.66 V    

3*(Cu2++2e−→Cu)    E∘=0.340 V

<u>Step 3:</u> The netto equation

2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-

2 Al + 3Cu2+  → 2Al3+ + 3Cu

<u>Step 4:</u> Calculate the standard cell potential

E∘cell = E∘cathode - E∘anode

E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al

E∘cell =0.340 V - (-1.66) = 2.00 V

The standard cell potential is 2.00 V

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