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Maslowich
3 years ago
5

A cylinder rod formed from silicon is 21.3 cm long and has a mass of 5.00 kg. The density of silicon is 2.33 g/cm3. What is the

diameter of the cylinder? (the volume of cylinder is given by π r2h, where r is the radius and h is the length)
Chemistry
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

11.32cm

Explanation:

This question describes a cylinder rod formed from silicon with length 21.3cm and mass 5kg. The density of the silicon is 2.33g/cm3.

To calculate the diameter, the radius is needed. To calculate the radius, the volume is needed. To calculate the volume, the formula: density = mass (m) / volume (V) is used.

Mass = 5kg, which is 5 kg × 1000g = 5000g

Density= 2.33g/cm3

Hence; volume= mass / density

= 5000/2.33

= 2145.9 cm3

Volume of cylinder= πr^2h

Where h= 21.3cm and π= 3.142

That is; r^2 = volume/πh

= r^2 = 2145.9/3.142×21.3

= r^2 = 2145.9/66.9246

= r^2 = 32.06

r= √32.06

r= 5.66cm

If radius of the cylinder is 5.66cm, the diameter is twice of the radius.

That is, diameter (d) = 5.66 × 2

= 11.32 cm

Therefore, the diameter of the cylinder is 11.32cm.

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Explanation:

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2) Manganesium

Mn^{2+}+2e^{-}---\ \textgreater \ Mn

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

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Add the two half-equations:

2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

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As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.
5 0
3 years ago
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Answer:

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Explanation:

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