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2 years ago

The ionization energies of Calcium are (in KJ/mol): 1st: 589.8 2nd: 1145.4 3rd: 4912.4

Chemistry

2 answers:

defon2 years ago

8 0

Answer:

See whole explanation to understand

Explanation:

the reason why there is such a large jump from 2nd to 3rd ionization energy for calcium is because to remove the third electron, a larger amount of energy is required, since the shell is closer to the nucleus, and higher attraction exists between them. This is why the second ionization energy is 1125.4 and then the third IE is 4912.4 which is a very big difference. It's all about the elections and energy!!

saw5 [17]2 years ago

4 0

Answer:

The ionization peak

Explanation:

i believe its due to the electron affinity, although im not sure.

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KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid

kari74 [83]

Explanation:

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

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3 years ago

What happens to the radius of non metal atom when it forms an ion?

MArishka [77]

Decrease because loss of electrons.

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2 years ago

What is 5.2034 x 10^8 in standard format

777dan777 [17]

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3 years ago

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An organic compound, which has the empirical formula C12H25 has an approximate molar mass of 338 g/mol. What is its probable mol

Marta_Voda [28]

Answer:

C24H50

Explanation:

The empirical fomula's molar mass is 169.25 g/mol.

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3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

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3 years ago