Answer:
i think mix im um 1s 6f 4d is correct answer
Answer:
B
Explanation:
First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.
Answer:
The energy released as heat when 9.94 g Cu 2 O ( s ) undergo oxidation at constant pressure is -10.142 kJ
Explanation:
Here we have
2Cu₂O ( s ) + O₂ ( g ) ⟶ 4 CuO ( s ) Δ H ∘ rxn = − 292.0 kJ mol
In the above reaction, 2 Moles of Cu₂O (copper (I) oxide) react with one mole of O₂ to produce 4 moles of CuO, with the release of − 292.0 kJ/mol of energy
Therefore,
1 Moles of Cu₂O (copper (I) oxide) react with 0.5 mole of O₂ to produce 2 moles of CuO, with the release of − 146.0 kJ of energy
We have 9.94 g of Cu₂O with molar mass given as 143.09 g/mol
Hence the number of moles in 9.94 g of Cu₂O is given as
9.94/143.09 = 6.95 × 10⁻² moles of Cu₂O
6.95 × 10⁻² moles of Cu₂O will therefore produce 6.95 × 10⁻² × − 146.0 kJ mol or -10.142 kJ.
<span>At 20g with half-life of 5 minutes there would be 10g leftover in 5 minutes. In 10 mins, there would be 5g leftover. Half-life is actually what it says, it refers to time it takes for half of the active elements, etc to break down at which time the potency of the "product" is half as strong. This term is used mainly with radioactive items or medicines.</span>
