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3 years ago

For each of the following sublevels, give the n and l values and the number of orbitals: (a) 5s; (b) 3p; (c) 4f

Chemistry

1 answer:

OverLord2011 [107]3 years ago

6 0

Answer:

(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1

(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Sublevel number, 0 ≤ l ≤ n − 1

So,

(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1

(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3

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Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

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3 years ago

Which statement identifies the element arsenic?

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Answer : Option A) Atomic number of Arsenic is 33.

Explanation : Arsenic contains same number of protons in its atomic nucleus. In arsenic there are 33 protons found in the atomic nucleus. Hence, the atomic number will be 33. It has 5 valence electrons in its outermost shell, which is also called as valence shell. So, its valency becomes 5.

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3 years ago

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

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Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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