S and S²⁻ do not have the outer subshell fully filled with electrons.
Explanation:
We look at electronic configurations:
Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons
S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons
Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons
S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons
Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons
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electron configurations
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Answer:
V = 34430 mL
Explanation:
Given data:
Volume in mL = ?
Number of moles of gas = 2.00 mol
Temperature = 36°C (36+273= 309K)
Pressure of gas = 1120 torr
Solution:
Formula:
PV = nRT
V = nRT/P
V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr
V = 38563.2 torr • L / 1120 torr
V = 34.43 L
L to mL
34.43 L ×1000 mL / 1 L
34430 mL
6 because after the 'H' is the number '6'