The term Reuther Ford gave to the positively charged particles in the nucleus of an antom was
1 answer:
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Answer:
(a) AL
(b) Sc
(c)Al
Explanation:
Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.
The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.
By looking at electron configuration we can figure out which electron will need more energy.
<h3>(a)Na, Mg, Al</h3>1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹
Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
<h3>(b) K, Ca, Sc</h3>K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
<h3>(c) Li, Al, B</h3>Li₃ ⇒ 1s², 2s¹
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
B₅ ⇒ 1s², 2s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
2) Determine the limitting reagent
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation: