The number of atoms present in 4-25 grams of NH 3 is pproximatelyA- 1 - 1023B- 1-5 - 1023C- 2 - 1023 NH 3D- 6-02 - 1023

The correct option is D 6.02 X 1023 4.25/17 × 4 × 6.023 × 10^23 = 6.023 × 10^23 ...

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The number of atoms present in $4.25 g r a m s$ of $N H_{3}$ is pproximately

1 X 10²³

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1.5 X 10²³

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2 X 10²³ NH3

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6.02 X 10²³

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The number of atoms present in $4.25 g r a m s$ of $N H_{3}$ is approximately

The number of molecules present in $4.25 g r a m s$ of $N H_{3}$ is approximately:

The number of atoms present in $4.25 g r a m s$ of $N H_{3}$ is approximately________.

The number of atoms present in 4.25 g of $({NH}_{3})$ is approximately.

a) $1 \times 10^{23}$

b) $1.5 \times 10^{23}$

c) $0.25 \times 10^{23}$

d) $6.02 \times 10^{23}$

THe number of molecules in 4.25 g of ammonia is:
(a) $1.0 \times 10^{23}$ (b) $1.5 \times 10^{23}$
(c) $2.0 \times 10^{23}$ (d) $3.5 \times 10^{23}$

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