Question

Calculate the no. Of barium and chloride ions present 104.1 g of bacl2

Answer 1

hey mate here is your answer..

Answer:Number Ba^{2+}Ba

2+

ions:=3.0103\times 10^{23} ions3.0103×10

23

ions

Number of Cl^-Cl

−

ions:=6.0106\times 10^{23} ions6.0106×10

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

208.23g/mol

104.1g

=0.4999mol

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

Number Ba^{2+}Ba

2+

ions:

1\times 3.0103\times 10^{23} ions=3.0103\times 10^{23} ions1×3.0103×10

23

ions=3.0103×10

23

ions

Number of Cl^-Cl

−

ions:

2\times 3.0103\times 10^{23} ions=6.0106\times 10^{23} ions2×3.0103×10

23

ions=6.0106×1

hope it may help u

Answer 2

Answer: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