Question

• No. of electron present in 3.6 mg of NH4+

Answer 1

18g of aNH4† contains NA no.

so, 3.6 × 10-³ contains X

So,

X = 6 × 10²³ × 3.6 × 10-³

———————————

18

X = 1.2 × 10*²°

For 3.6 mg of NH4+ ,

No. Of electrons

1.2 × 10²° × 10

= 1.2 × 10²¹

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Answer 2

Answer:$12.046$ ×$10^{20} electrons$ are present in $3.6mg$ of NH4+.

Explanation:

• Given mass of NH4+ = $3.6mg$

                                            = $3.6$×$10^{-3}$$g$

• Molar mass of NH4+ $=$( $1$× $14$ + $4$×$1$ )

                                           $=14 + 4\\= 18g$

• Number of moles =  Given mass/ Molar Mass

                                       $=\frac{0.0036}{18} \\= 2 * 10^{-4}$ moles of NH4+.

1 mole of NH4+ = $6.023 * 10^{23} electrons$

$0.0002$ moles of NH4+  $=0.0002* 6.023*10^{23} *10$

                                      $=12.046 * 10^{20} electrons$

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