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Veseljchak [2.6K]
3 years ago
10

If Na was to form a 2 ion, from what orbital subshell would the second electron be lost?

Chemistry
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

2p

Explanation:

Sodium has an atomic number of 11, thus, the neutral atom has 11 electrons. The electron configuration is:

Na: 1s² 2s² 2p⁶ 3s¹

To gain stability, it loses an electron from its outer shell to form the cation Na⁺. Its electron configuration is:

Na⁺: 1s² 2s² 2p⁶

If it were to lose a second electron to form a Na²⁺ cation, the electron should be lost from the 2p orbital subshell and its electron configuration would be:

Na²⁺: 1s² 2s² 2p⁵

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A compound composed of nickel and fluorine contains 9.11 g ni and 5.89 g f. what is the empirical formula of this compound
aalyn [17]
Empirical formula is the simplest formula showing the simplest ratio of atoms in a compound. Calculated as shown;
 we start by calculating the number of moles of each atom;
moles of nickel = 9.11 g ÷ 58.7 g = 0.155 moles 
moles of fluorine = 5.89 g ÷ 19 g = 0.31 moles
Then we get the ratio of the moles of nickel to that of flourine
 That is 0.155 : 0.31  (dividing by the smallest)
            0.155/0.155 : 0.31/0.155
we get 1:2  ( the simplest ratio)
Therefore the empirical formula is nif2 
6 0
3 years ago
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Tju [1.3M]

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Complex compound (

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4 0
3 years ago
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choli [55]

Answer:

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Explanation:

7 0
3 years ago
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Anna [14]

Answer:

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3 0
3 years ago
Consider the reaction mg(oh)2(s)→mgo(s)+h2o(l) with enthalpy of reaction δhrxn∘=37.5kj/mol what is the enthalpy of formation of
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ΔHf(Mg(OH)₂) = <span>−924,5 kJ/mol.
</span>ΔHf(H₂O) = <span>−285,8 kJ/mol.
</span>ΔHrxn -enthalpy of reaction.
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ΔHf(MgO) = -601,2 kJ/mol.
7 0
3 years ago
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