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1 year ago

At standard temperature and pressure. 0.500 mole of xenon gas occupies

Chemistry

1 answer:

ANEK [815]1 year ago

3 0

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

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Consider the SCl2 molecule. (a) What is the electron config- uration of an isolated S atom? (b) What is the electron con- figura

iragen [17]

Answer:

(a) 1s² 2s² 2p⁶ 3s² 3p⁴

(b) 1s² 2s² 2p⁶ 3s² 3p⁵

(d) No valence orbital remains unhybridized.

Explanation:

Consider the SCl₂ molecule.

(a) What is the electron configuration of an isolated S atom?

S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.

(b) What is the electron configuration of an isolated Cl atom?

Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.

(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂?

SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.

(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?

No valence orbital remains unhybridized.

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3 years ago

Which of the following substances would likely dissolve but not dissociate in water

AlexFokin [52]

Answer:

Oils, fats, and certain organic solvents do not dissolve in water because they are hydrophobic.

Explanation:

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2 years ago

Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t

Dvinal [7]

Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % 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 Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

$6.941=[(6.01512\times x)+(7.01600\times (100-x))]$

By solving the term 'x', we get:

$x=694.048$

Percent abundance of Li isotope-1 = $\frac{694.048}{100}=6.94\%$

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

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2 years ago

Pls help —————————— —————-

Marina86 [1]

Answer:

X has 9 Electrons, 9 Protons and 10 Neutrons

X is Fluorine

Explanation:

19 in X is Nucleon number

So, if you minus 19 by 9 down there (which is proton number) you'll get 10 (Neutron number).

Now, Proton = Electron. That's why Electron is 9 too. I figured that X is Fluorine bcuz it has 9 Electrons. If you venture around in the periodic table, fluorine is the ninth element innit.

hope this helps ya ＼(^o^)／

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2 years ago

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2 years ago