Question

(a) Show the formula of Sodium carbonate using criss cross method
(b) Choose the cation and anion present in CH3COONa.
(c) Write the difference between O2, and 2O.
(d) Calculate the number of atoms of Helium in
i) 52g of Helium.
i) 52u of Helium​

Answer 1

Answer:

A. The chemical formula of sodium carbonate in criss cross method is . Explanation: For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Explanation:

D. 1.The number of atoms in 52g of He is-

78.299×1024 atoms.

7.820×10-24 atoms.

7.829×1024 atoms.

78.234×12025 atoms.

D. 2.=7.8286 × 10^24. In 52 u of He 7.8286 × 10^24 atoms are present