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Postsecondary • 3mo.

Hi there!

I need explanation for question 5 and 6 as soon as possible because tomorrow is my exam and I’m not able to understand how to answer them. If someone could kindly explain me in details in steps.

The answers are below the questions but I don’t get how to solve them?

thank u so much!



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Explanations (2)

  • Explanation from Alloprof

    Explanation from Alloprof

    This Explanation was submitted by a member of the Alloprof team.

    Team Alloprof • 3mo.


    We're sorry we weren't able to answer your question.

    Next time, please take more time to allow us to answer your question.

    Also, we can't look at the whole problem, so you'll have to show us your procedure.

    Thank you very much. See you soon :)

  • Options
    3mo. edited September 2


    Density of gold d=19.3 g cm-³

    Area of the sheet S=124.5 ft²

    h thickness of the sheet to be determined

    m mass of gold used m=28.35 g

    we have the formula

    mass= density x Volume

    Volume = Area x thickness

    m=dxSxh h=m/d x S

    units given

    m : g

    d : g cm-³

    S: ft²

    we have to change S in cm²

    1 ft = 12 inch

    1 inch =2.54 cm

    we obtain

    1 ft = 12 x 25.4 cm 1ft=30.48 cm

    1ft²= 30.48² cm²

    1ft²= 929.0304 cm²

    S=(124.5 x 929.0305) cm²

    h= 28.35 g/(19.3 g cm-³ (124.5 x 929.0305) cm²)

    h=28.35 g/(19.3 g cm-³ x 124.5 x 929.0305 cm²)

    h=(28.35 /(19.3 x 124.5 x 929.0305 )) (g/g cm-³ cm²)

    h=(0.000012699) cm

    1 cm =0.01 m

    h=(0.000012699) x 0.01 m

    h=0.00000012699 m

    h=126.99 nm =127 nm


    atom gold diameter =288 pm

    thickness of the sheet=(number of atoms) x (atom gold diameter)

    other wise





    m = V x d

    mass = Volume x density

    28.35 g = V x 19.3 g cm-³

    V=28.35 g/19.3 g cm-³

    V=(28.35/19.3) g/g cm-³

    V=1.469 cm³

    we consider this volume as a cube



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