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ZanzabumX [31]
3 years ago
13

Write the ground state electron configuration of zn using the noble-gas shorthand notation.

Chemistry
1 answer:
V125BC [204]3 years ago
8 0
Atomic Number of Zinc is 30, means it contains 30 electrons. So, its electronic configuration is as follow,

                                    1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰
As,
          1s², 2s², 2p⁶, 3s², 3p⁶  =  Argon
So,
Electronic configuration of Zinc in shorthand notation is as follow,

                                     [Ar] 4s², 3d¹⁰
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A gas has a volume of 5.00 L at 0°C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to ea
Diano4ka-milaya [45]

Answer:

A = -213.09°C

B = 15014.85 °C

C = -268.37°C

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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