Chem Help!!

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A standard 1.000kg- mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 3.00in. The density of the steel is 7.70g/cm^3.


How many inches long must the section of bar be?

  • Chem Help!! -

    mass=density(volume)=density*area*length

    where area=sqrt(s(s-a)^3) s=half perimeter=(3in*2.54cm/in)*3/2
    area=sqrt(9*1.27(3*2.54*2))

    check that.

  • Chem Help!! -

    thank you

  • Chem Help!! -

    How many inches is the section of the bar

  • Chem Help!! -

    I'm not sure that I understand what you did, but this is what I did.


    1kg=1 x10^3 g

    Density=Mass/volume, so

    Volume =mass/density= (1 x10^3 g/7.70g/cm^3)= 130cm^3

    Since in a equilateral triangle all sides are equal and since volume = cm*cm*cm or l*w*h

    (130cm^3)^(1/3)=5.06cm

    Converting to inches
    5.06cm*(1in/2.54)= 2 inches/side

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