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Maurinko [17]
3 years ago
9

Sort these species into isoelectronic groups. It doesn\'t matter which group goes in which box, so long as the correct species a

re grouped.
He
Be2+
Li+
N3-
Ne
S2-
Mg2+
Ti4+
K+
Chemistry
1 answer:
OleMash [197]3 years ago
7 0

Answer : The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

S^{2-},Ti^{4+},K^{+}

Explanation :

Isoelectronic : It is defined as the compound or molecule having the same number of electrons and the same number of electronic structure.

  • The element is helium. The number of electrons are 2.
  • The element is beryllium. The number of electrons are 4. The number of electrons in Be^{2+} = 4 - 2 = 2
  • The element is lithium. The number of electrons are 3. The number of electrons in Li^{+} = 3 - 1 = 2
  • The element is nitrogen. The number of electrons are 7. The number of electrons in N^{3-} = 7 + 3 = 10
  • The element is neon. The number of electrons are 10.
  • The element is sulfur. The number of electrons are 16. The number of electrons in S^{2-} = 16 + 2 = 18
  • The element is magnesium. The number of electrons are 12. The number of electrons in Mg^{2+} = 12 - 2 = 10
  • The element is titanium. The number of electrons are 22. The number of electrons in Ti^{4+} = 22 - 4 = 18
  • The element is potassium. The number of electrons are 19. The number of electrons in K^{+} = 19 - 1 = 18

The isoelectronic groups are:

He,Be^{2+},Li^{+}

N^{3-},Mg^{2+}

S^{2-},Ti^{4+},K^{+}

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a solution is made by dissolving 14.57 g of sodium bromide nabr in 415 of water . what is the molality of the solution?
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Explanation:

Data

mass of NaBr = 14.57 g

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2.- Calculate the moles of NaBr

                     103 g of NaBr ------------------ 1 mol

                      14.57 g of NaBr ---------------  x

                               x = (14.57 x 1) / 103

                               x = 14.57 / 103

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