1. (a) Molecular mass of glucose (C₆H₁₂O₆) = 6 x 12 + 12x1 + 6 x 16 = 72 + 12 + 96 = 180 u. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/

    (a) Molecular mass of glucose (C₆H₁₂O₆) = 6 x 12 + 12×1 + 6 x 16 = 72 + 12 + 96 = 180 u.

    https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/

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  2. (C) Mg + S → MgS 24 g 32 g Ratio by mass in which Mg and S combine = 24:32 or 3:4 https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/

    (C) Mg + S → MgS
    24 g 32 g
    Ratio by mass in which Mg and S combine = 24:32 or 3:4

    https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-3/

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  3. (c) Changes (i), (iii) and (iv) are physical in nature because we can get back to the original form by physical methods. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-2/

    (c) Changes (i), (iii) and (iv) are physical in nature because we can get back to the original form by physical methods.

    https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-2/

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