Question

Calculate the no. Of Ba+2 ions and Cl-ions present in 104.1g of anhydrous BaCl2

Answer 1

Number Ba^{2+} ions:=3.0103\times 10^{23} ions

Number of Cl^- ions:=6.0106\times 10^{23} ions

Explanation:

Mass of barium chloride = 104.1 g

Mole of barium chloride = \frac{104.1 g}{208.23 g/mol}=0.4999 mol

Number molecules of barium chloride =0.4999 \times 6.022\times 10^{23]=3.0103\times 10^{23} molecules

Number Ba^{2+} ions:

1\times 3.0103\times 10^{23} ions=3.0103\times 10^{23} ions

Number of Cl^- ions:

2\times 3.0103\times 10^{23} ions=6.0106\times 10^{23} ions

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Answer 2

Answer:

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