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3 years ago

How many electrons in an atom can have each of the following quantum number or sublevel designations?

Chemistry

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Answer: (a) 2 (b) 6 (c) 14

Explanation:

In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).

s is l=0, p is l=1, d is l=2, f is l=3.

The magnetic quantum number (ml) The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.

(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0 accommodate two(2)electrons

(b) 5p. p is l=1 If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.

(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.

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3 years ago

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What is the [h3o+] in a solution that consists of 1.0 m nh3 and 2.5 m nh4cl? [kb (nh3) = 1.8 ×10-5]a) 1.1 × 10-5

Archy [21]

Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?

Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

$\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}$ .

The $\text{K}_b$ value for ammonia is small. The value of x will be so small that at equilibrium, $1.0 - x \approx 1.0$ and $2.5- x \approx 2.5$ .

$\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}$ .

$\displaystyle \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0} = 1.8\times 10^{-5}$ .

$\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}$ .

Again, $\text{K}_w = 1.0\times 10^{-14}$ at room temperature.

$\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}$

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3 years ago

You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H

KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+ pKa1 = 2.14

H2PO4 = HPO4 + H+ pKa2 = 6.86

HPO4 = PO4 + H+ pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

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3 years ago

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Ass

yaroslaw [1]

Answer:

moles of ammonia produced = 0.28 moles

Explanation:

The reaction is

$N_{2}(g)+3H_{2}(g) --> 2NH_{3}(g)$

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia

So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen

= 0.42 moles of hydrogen molecule.

this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia

the moles of ammonia produced = 0.28 moles

Here the nitrogen is limiting reagent.

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3 years ago

Which two systems first respond to burning your finger?

Bess [88]

Answer:

B. is my guess

I'm sorry if I'm wrong

God bless :)

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