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1 year ago

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There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 12.01%. Now that you have the contribution

from the X-19 isotope (2.282) and from the X-21 isotope (18.48), what is the average atomic mass (in amu) of this element using four significant figures

1 answer:

mestny [16]1 year ago

4 0

Answer: The average atomic mass of X is 16.53

Explanation:

Mass of isotope X-19 = 2.282

% abundance of isotope X-19 = 12.01% = $\frac{12.01}{100}=0.1201$

Mass of isotope X-21 = 18.48

% abundance of isotope X-21 = (100-12.01)% = $\frac{100-12.01}{100}=0.8799$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(2.282\times 0.1201)+(18.48\times 0.8799)]$

$A=16.53$

Therefore, the average atomic mass of X is 16.53

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How many values of ml are allowed for an electron in a 5f subshell??

VladimirAG [237]

There are 7 values of ml are allowed for an electron in a 5f subshell namely: -3 -2, -1, 0, +1, +2, +3

<h3>Further explanation</h3>

In an atom, there are energy levels in the shell and subshell.This energy level is expressed in terms of electron configurations.

Writing the electron configuration starts from the lowest to the highest subshell's energy level. There are 4 sub-shells in an atom's shell, namely s, p, d, and f. The maximum number of electrons for each subshell is

s: 2 electrons
p: 6 electrons
d: 10 electrons and
f: 14 electrons

Electron filling in subshells using the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Each sub-shell also has orbitals drawn in the form of a square box in which there are electrons symbolized by half arrows.

Each orbital in an atom consists of 4 quantum numbers

n is the principal quantum number.
l is the angular momentum / azimuthal quantum number
ml, the magnetic quantum number
ms, the electron-spin quantum number

Value of n: positive integer

value of l: s = 0, p = 1, d = 2, f = 3

ml value: between -l to + l

ms value: +1/2 or -1/2

Determination of electron configurations based on principles:

1. Aufbau: Electrons occupy orbitals of the lowest energy level
2 Hund: electron fills orbitals with the same energy level
3. Pauli: there are no electrons that have 4 equal quantum numbers

So for 5f orbitals, the value of a possible quantum number is

n = 5;

l = 3 (f = 3);

m = -3 -2, -1, 0, +1, +2, +3;

s = + - 1/2

<h3>Learn more</h3>

the locations and properties of two electrons

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Keywords: orbitals, subshells, quantum numbers

8 0

1 year ago

Read 2 more answers

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

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2 years ago

consider the reaction between sulfite and a metal anion, X2-, to form the metal, X, and thiosulfate: 2 X2-(aq) + 2 SO32- + 3 H2O

Dafna11 [192]

Answer:

$E_{red}^{0}$ for X is -1.20 V

Explanation:

Oxidation: $2\times$ [ $X^{2-}(aq.)-2e^{-}\rightarrow X(s)$ ]

reduction: $2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

---------------------------------------------------------------------------------------------------

overall: $2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

So, $E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})$

or, $0.63=-0.57-E_{red}^{0}(X\mid X^{2-})$

or, $E_{red}^{0}(X\mid X^{2-})= -1.20$

So, $E_{red}^{0}$ for X is -1.20 V

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2 years ago

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A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

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2 years ago

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yKpoI14uk [10]

B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.

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