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mariarad [96]
3 years ago
8

A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.

Chemistry
1 answer:
adell [148]3 years ago
8 0

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

c) \rho=1.5x10^5g/L

Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

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