Explain why NaBH4 is less reactive than LiAlH4 (Hint: write the structure of each ion, (BH4) - and (AlH4) - , with their formal charges

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Explanation:

The ability of a hydride to act as a reducing agent depends on the electro negativity of the central atom in the complex hydride anion.

If we compare the electronegativities of Boron and Aluminum, Boron is more electronegative than aluminum. The more electronegative the central atom is, the lesser the electron density on the hydride and the lesser its ability to act as a reducing agent.

For the reason stated above, LiAlH4, having aluminium of low electro negativity as its central atom is a better reducing agent than NaBH4.

sry but any answer choices

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%