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Serjik [45]
2 years ago
9

What is the density of nitrogen gas at 356.4 K?

Chemistry
1 answer:
Ugo [173]2 years ago
4 0

Answer:

Nitrogen gas (chemical symbol N) is generally inert, nonmetallic, colorless, odorless and tasteless. Its atomic number is 7, and it has an atomic weight of 14.0067. Nitrogen has a density of 1.251 grams/liter at 0 C and a specific gravity of 0.96737, making it slightly lighter than air.

Explanation:

:)

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Compare and contrast all the periodic trends as you move down a group of metais. Organize
elena55 [62]

Answer:

The number of energy levels increases as you move down a group as the number of electrons increases. Each subsequent energy level is further from the nucleus than the last. Therefore, the atomic radius increases as the group and energy levels increase.

Explanation:

Dose this help? Tell me if it dose.

4 0
3 years ago
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What is the easiest way for atoms without a full valence shell to gain a full valence shell?
sergey [27]

Most atoms do not. For those atoms that do not have a full valence shell (which usually would contain eight electrons, except for hydrogen and helium, where it would contain two), something has to change. So nature's tendency toward a full valence shell will lead to one of two things: The gain or loss of electrons.

7 0
3 years ago
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Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
Name the reaction type & mechanism of the following reactions:
Natali5045456 [20]

Answer:

MECHANISM:

1) The lone pair on oxygen attacks the H-Br molecule forming a hydronium ion.

2) Formation of carbocation.

3) Attack of Nucleophile Br  −  .

Explanation:

4 0
2 years ago
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Can some atoms exceed the limits of the octet rule in bonding? If so, give an example.
harkovskaia [24]

Answer:

Yes. Example: <u>Sulfur hexafluoride (SF₆) molecule</u>

Explanation:

According to the octet rule, elements tend to form chemical bonds in order to have <u>8 electrons in their valence shell</u> and gain the stable s²p⁶ electronic configuration.

However, this rule is generally followed by main group elements only.

Exception: <u>SF₆ molecule</u>

In this molecule, six fluorine atoms are attached to the central sulfur atom by single covalent bonds.

<u>Each fluorine atom has 8 electrons in their valence shells</u>. Thus, it <u>follows the octet rule.</u>

Whereas, there are <u>12 electrons around the central sulfur atom</u> in the SF₆ molecule. Therefore, <u>sulfur does not follow the octet rule.</u>

<u>Therefore, the SF₆ molecule is known as a </u><u>hypervalent molecule</u><u> or expanded-valence molecule.</u>

6 0
3 years ago
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