Atoms present in a molecule of heavy water are:

_{1} H^{1},_{8} O^{16}

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_{1} H^{2},_{8} O^{18}

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_{1} H^{2},_{8} O^{16}

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_{1} H^{1},_{8} O^{18}

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Solution

The correct option is C $_{1} H^{2},_{8} O^{16}$
The formula of heavy water is $D_{2} O$ . $D (_{1} H^{2})$ is an isotope of hydrogen.

Atoms present in a molecule of heavy water are $_{1} H^{2}$ and $_{8} O^{16}$ i.e., deuterium $(_{1} H^{2})$ and normal oxygen $(_{8} O^{16})$ .

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Ordinary water contains one part of heavy water per 6000 parts by weight.The number of heavy water molecules present in a drop of water of volume 0.01 ml is (Density of water is 1 g/ml)

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Ordinary water contains one part of heavy water per 6000 parts by weight.The number of heavy water molecules present in a drop of water of volume 0.01 ml is (Density of water is 1 g/ml)

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Atoms present in a molecule of heavy water are:

The correct option is C 1H2,8O16The formula of heavy water is D2O. D (1H2) is an isotope of hydrogen.Atoms present in a molecule of heavy water are 1H2 and 8O16 i.e., deuterium (1H2) and normal oxygen (8O16).

The correct option is C $_{1} H^{2},_{8} O^{16}$
The formula of heavy water is $D_{2} O$ . $D (_{1} H^{2})$ is an isotope of hydrogen.

Atoms present in a molecule of heavy water are $_{1} H^{2}$ and $_{8} O^{16}$ i.e., deuterium $(_{1} H^{2})$ and normal oxygen $(_{8} O^{16})$ .