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Masja [62]
3 years ago
15

The first two ionization energies of nickel.

Chemistry
1 answer:
Radda [10]3 years ago
7 0
The first two ionization energy of nickel are expressed as:
Ni(g) = Ni^+(g) + e^- Ni^+(g) = Ni^2+(g) + e^- 
IOnization energy os the energy that is neccessary in order to remove one mole of electrons from an atom.
Hope this helps.
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An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g
vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= \frac{33.8 g solute}{100 g solution} x \frac{1.15 g solution}{1 ml} x \frac{1 mol solute}{145.6 g solute} x \frac{1000 ml}{1 L}

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x \frac{1 mol solute}{145.6 g} = 0.232 mol

mol H20= 66.2 g H₂O x \frac{1 mol H2O}{18 g}

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute=\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}

Xsolute= 0.059

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What is the frequency of electromagnetic radiation having a wavelength of 3.27 ✕ 10-8 m? s-1 What type of electromagnetic radiat
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PLEASE HELP ME The pH scale for acids and bases ranges from _____.
inysia [295]

The pH scale for acids and bases ranges from 1 - 14. The answer is letter C. The rest of the choices do not answered the question above. There are quite a few relationships between [H+] and [OH−] ions. And because there is a large range of number between 10 to 10-15 M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, [H+ ][OH- ] = 10-14. From here we can derive the values of each concentration.

 


 -log[H+ ] + -log[OH- ] = -log[10-14]


pH + pOH = 14


So pH = 14 – pOH and pOH = 14 –

pH 


It would be B.

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3 years ago
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