Question

From 280mg of co 10power21 molecules are removed calculate no. Of moles of CO left​

Answer 1

Answer:

∵6.023×10^23 molecules of CO=28g

$∴ {10}^{21} molecules \: of \: co = \frac{28 \times {10}^{21} }{6.023 \times {10}^{23} } = 4.65 \times {10}^{ - 2} g = 46.5 \: mg$

∴ CO left =280−46.5=233.5mg

$also \: moles \: of \: co \: left = \: \frac{wt.}{m.wt.} =$

$\frac{233.5 \times {10}^{ - 3} }{28} = 8.34 \times {10}^{ - 3}$

Answer 2

Answer:

CO.280mg=

co⋅10=10co

21=10co

CO=10co

Explanation: