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saul85 [17]
4 years ago
6

In the process of ionization, what is the relationship between the second ionization energy (I2) and the third ionization energy

(I3)?
Chemistry
1 answer:
dimulka [17.4K]4 years ago
4 0
<span>The first ionization energy is the energy that is obligatory in order to eliminate the first electron from an atom(neutral) in the GAS phase
, the second ionization energy is defined as  the energy required to remove the second electron from an atom(+1 ion) in the GAS phase.
</span>the third ionization energy is the energy it takes to remove an electron from a 2+ ion. (That means that the atom has already lost two electrons, you are now removing the third.) 
<span> Ionization energy will normally increase for every electron that is removed and increases from left to right in the periodic table and moving up the periods
hope it helps</span>
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The orbit closest to the nucleus has ___________ energy.
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inonic bonds with cavalent bonds

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ionic bonds

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An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
seraphim [82]

Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
4 years ago
What is the molar mass of a substance
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4 years ago
Write the balanced chemical equation for the complete, stoichiometric combustion of ethylene in (a) nitrous oxide and (b) air. C
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<u>Answer:</u> Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.

<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 a:</u> Combustion of ethylene in nitrous oxide

The chemical equation for the combustion of ethylene in nitrous oxide follows the equation:

C_2H_4+N_2O\rightarrow 2CO_2+2H_2O+6N_2

By stoichiometry of the reaction;

6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.

Calculating the mass of nitrous oxide using equation 1:

Molar mass of nitrous oxide = 44 g/mol

Moles of nitrous oxide = 6 moles

Putting values in equation 1, we get:

6mol=\frac{\text{Mass of nitrous oxide}}{44g/mol}\\\\\text{Mass of nitrous oxide}=264g

Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.

  • <u>For b:</u> Combustion of ethylene in air

The chemical equation for the combustion of ethylene in air follows the equation:

C_2H_4+3O_2\rightarrow 2CO_2+2H_2O

By stoichiometry of the reaction;

3 moles of oxygen are required for the complete combustion of ethylene molecule.

Calculating the mass of oxygen using equation 1:

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 3 moles

Putting values in equation 1, we get:

3mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=96g

Thus, 96 grams of oxygen are required for the complete combustion of ethylene.

7 0
3 years ago
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