Question

FIND OUT THE ELECTRONIC CONFIGURATION. Helium (Z=2, A=4) 2. Boron (Z=5, A =11) 3. Oxygen (Z=8, A=16) 4. Silicon ( Z=14, A =28) 5. Argon (Z=18, A= 40)
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Answer 1

Explanation:

II.C.12 Hydrogen Fluoride–Tantalum Pentafluoride

HF:TaF5 is a catalyst for various hydrocarbon conversions of practical importance. In contrast to antimony pentafluoride, tantalum pentafluoride is stable in a reducing environment. The HF:TaF5 superacid system has attracted attention mainly through the studies concerning alkane alkylation and aromatic protonation. Generally, heterogeneous mixtures such as 10:1 and 30:1 HF:TaF5 have been used because of the low solubility of TaF5 in HF (0.9% at 19 °C and 0.6% at 0 °C). For this reason, acidity measurements have been limited to very dilute solutions, and an H0 value of −18.85 has been found for the 0.6% solution. Both electrochemical studies and aromatic protonation studies indicate that the HF:TaF5 system is a weaker superacid than HF:SbF5.

Answer 2

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