1 and 5 are definitely significant. The zeros might not be significant .
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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H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>
Answer:
Equation: - 2Li + H2O = Li2O + H2 Uncoated lithium metal reacts with water to form a colorless lithium hydroxide solution and hydrogen gas.
Explanation:
The ideal gas law may be written as

where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)