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Srushti

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  1. Asked: December 23, 2023In: Class 10 Science

    The image formed by a concave mirror is observed to be real, inverted and larger than the object. Where is the object placed?

    Srushti
    Added an answer on December 23, 2023 at 10:34 am

    Between the principal focus and the centre of curvature. When the image formed by a concave mirror is real, inverted, and larger than the object, the object must be located beyond the focal point (F) of the mirror. In this case, the object is positioned between the focal point (F) and the mirror's cRead more

    Between the principal focus and the centre of curvature. When the image formed by a concave mirror is real, inverted, and larger than the object, the object must be located beyond the focal point (F) of the mirror. In this case, the object is positioned between the focal point (F) and the mirror’s center of curvature (C).

    To summarize:
    • Image is real: A real image is formed when the reflected rays actually converge, and it can be projected onto a screen.
    • Image is inverted: The orientation of the image is upside down compared to the object.
    • Image is larger than the object: The magnification is greater than 1, resulting in an enlarged image.
    • So, for a concave mirror with a real, inverted, and larger image, the object is placed beyond the focal point but inside the center of curvature.

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  2. Asked: December 23, 2023In: Class 10 Science

    Both a spherical mirror and a thin spherical lens have a focal length of (-)15 cm. What type of mirror and lens are these?

    Srushti
    Added an answer on December 23, 2023 at 10:34 am

    Both are concave. A concave mirror and a concave lens both have a focal length with a negative sign. Therefore, a spherical mirror with a focal length of (-)15 cm is a concave mirror. Similarly, a thin spherical lens with a focal length of (-)15 cm is a concave lens. In the context of mirrors and leRead more

    Both are concave. A concave mirror and a concave lens both have a focal length with a negative sign. Therefore, a spherical mirror with a focal length of (-)15 cm is a concave mirror. Similarly, a thin spherical lens with a focal length of (-)15 cm is a concave lens.
    In the context of mirrors and lenses:

    • For mirrors, positive focal lengths are associated with convex mirrors, while negative focal lengths are associated with concave mirrors.

    • For lenses, positive focal lengths are associated with converging lenses (convex lenses), while negative focal lengths are associated with diverging lenses (concave lenses).

    • Alternative, answer that should be given credit: Plano-concave lens

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  3. Asked: December 23, 2023In: Class 10 Science

    Identify the products formed when 1 mL of dil. Hydrochloric acid is added to 1g of Sodium metal.

    Srushti
    Added an answer on December 23, 2023 at 10:34 am

    When dilute hydrochloric acid (HCl) is added to sodium metal (Na), a chemical reaction takes place, resulting in the formation of sodium chloride (NaCl) and the liberation of hydrogen gas (H2). The balanced chemical equation for the reaction is: 2Na+HCl →2NaCl+H2 So, for the reaction of 1 mL of diluRead more

    When dilute hydrochloric acid (HCl) is added to sodium metal (Na), a chemical reaction takes place, resulting in the formation of sodium chloride (NaCl) and the liberation of hydrogen gas (H2).
    The balanced chemical equation for the reaction is:

    2Na+HCl →2NaCl+H2

    So, for the reaction of 1 mL of dilute hydrochloric acid with 1 g of sodium metal, you can expect the formation of sodium chloride and the evolution of hydrogen gas. The balanced equation indicates that two moles of hydrochloric acid react with two moles of sodium to produce two moles of sodium chloride and one mole of hydrogen gas. If you’re working with a specific amount of sodium (1 g in this case), you can use the molar mass to determine the moles of sodium and then apply the stoichiometry of the balanced equation to find the expected amounts of sodium chloride and hydrogen gas formed.

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  4. Asked: December 23, 2023In: Class 10 Science

    Write the chemical name and chemical formula of the salt used to remove permanent hardness of water.

    Srushti
    Added an answer on December 23, 2023 at 10:34 am

    Sodium carbonate decahydrate has the chemical formula Na2CO3.10H2O, indicating that each formula unit of sodium carbonate is associated with 10 water molecules. This compound is commonly known as sodium carbonate decahydrate or sodium carbonate decahydrate.

    Sodium carbonate decahydrate has the chemical formula Na2CO3.10H2O, indicating that each formula unit of sodium carbonate is associated with 10 water molecules. This compound is commonly known as sodium carbonate decahydrate or sodium carbonate decahydrate.

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  5. Asked: December 23, 2023In: Class 10 Science

    List any two observations when Ferrous Sulphate (FeSO₄) is heated in a dry test tube?

    Srushti
    Added an answer on December 23, 2023 at 10:34 am

    When ferrous sulfate (FeSO₄) is heated in a dry test tube, two observations can be made: Color Change: Initially, ferrous sulfate is typically green or bluish-green in color. As it is heated, the water of crystallization is driven off, and the color of the compound may change. The hydrated form of fRead more

    When ferrous sulfate (FeSO₄) is heated in a dry test tube, two observations can be made:

    Color Change: Initially, ferrous sulfate is typically green or bluish-green in color. As it is heated, the water of crystallization is driven off, and the color of the compound may change. The hydrated form of ferrous sulfate, known as iron(II) sulfate heptahydrate (FeSO₄·7H₂O), loses water molecules upon heating, and the color may change to white or a lighter shade.

    Formation of Oxides: As the temperature increases, ferrous sulfate undergoes thermal decomposition. This process leads to the formation of iron oxides, such as iron(II) oxide (FeO) or iron(III) oxide (Fe₂O₃), depending on the specific conditions and the extent of heating. The color change associated with the formation of these oxides can be observed, and in some cases, the residue may appear reddish-brown or black.

    Hope you like it……..👍👍👍

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