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Sav [38]
3 years ago
14

You have 2.394 x 102 mL of oxygen gas (O2) at 7.20 x 102 mm of Hg and 78 oC.

Chemistry
1 answer:
igomit [66]3 years ago
3 0
We are given with
V = <span>2.394 x 102 mL 
P = </span><span>7.20 x 102 mm of Hg
T = </span><span>78 oC

We are asked to determine the mass of the sample in milligrams
Using the ideal gas law
n = PV / RT
n = </span>7.20 x 102 mm of Hg (2.394 x 102 mL) / R (78 + 273)
Use the appropriate value for R or just convert the values to SI and use R = 8.314
Then, solve for the mass of the sample by using the MW Oxygen which is 16 g/mol
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The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO(g)
AVprozaik [17]

Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO(g) + NO₂(g) → N₂O₃(g) ΔH o rxn = −39.8 kJ

(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH o rxn = −112.5 kJ

(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

  • If the chemical equation is inverted, the symbol of ΔH is also reversed.
  • If the coefficients are multiplied, multiply ΔH by the same factor.
  • If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)  you must do the following:

  • Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

<em>4 NO₂(g) →  2 N₂O₄(g)  ΔH o rxn = −114.4 kJ</em>

  • Reverse equations (1) and (2)

(1) <em>N₂O₃(g)  → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ</em>

(2) <em>N₂O₅(g) →  NO(g) + NO₂(g) + O₂(g)  ΔH o rxn = 112.5 kJ</em>

Equations (4) and (5) are maintained as stated.

(4) <em>2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ </em>

(5) <em>N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ </em>

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

<u><em>The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ</em></u>

8 0
3 years ago
Which electron configuration is correct for a sodium ion?
jeka94

The electron configuration for sodium ion Na⁺ : <u>(2) 2-8</u>

<h3>Further explanation </h3>

In an atom there are levels of energy in the shell and sub shell

This energy level is expressed in the form of electron configurations.

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

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

Charging electrons in the sub shell uses the following sequence:

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc. </em>

Determination of electron configurations based on principles:

• 1. Aufbau: Electrons occupy orbitals of the lowest energy level

• 2 Hund: electrons fill orbitals with the same energy level

• 3. Pauli: no electrons have the same 4 quantum numbers

The alkali metal Na will release electrons to form Na + so that the electron configuration is stable as the noble gas element Ne

electron configuration Ne: [He] 2s² 2p⁶

Na electron configuration: [Ne] 3s¹

electron configuration Na + = [Ne] = [He] 2s² 2p⁶

The maximum number of electrons in the shell K, L, M, N

According to Bohr, the maximum number of electrons that can occupy each atomic shell can be calculated by the formula 2n²

  • K shell (n = 1): 2.1² = 2 electrons
  • L shell (n = 2): 2. 2² = 8 electrons
  • M shell (n = 3): 2. 3² = 18 electrons
  • N shell (n = 4): 2. 4² = 32 electrons

If we look at the configuration of the Na⁺ ion:

1s² 2s² 2p⁶ then

on the shell n=1 (1s) there are 2 electrons

on the shells n=2 (2s and 2p) there are 8 electrons

So that the configuration

Na⁺ = 2-8

<h3>Learn more </h3>

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Which sentence describes two atoms that form a covalent bond?
marin [14]

Answer: A. The electronegativity of one atom is less than 1.7

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How many moles of Fe2O3 are in 17.2g?<br><br> A - 1.23<br> B - 2.75<br> C - 0.239<br> D - 0.108
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Answer:

C

Explanation:

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Answer:

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