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3 years ago

A cylinder of lead has a radius of 4.0cm and a height of 10.0cm. What is the mass of the cylinder of lead?

Chemistry

1 answer:

Korolek [52]3 years ago

4 0

Answer:

Explanation:

So lead has a density of 11.34 g/cm^3 according to a quick google search. If this is not the value you are using, just replace what you do use where I put 11.34

knowing the density, if we knew the volume we could find the mass by unit conversion since g/cm^3 * cm^3 = g, so we would multiply the density by the volume. And the question gives us the means to find the volume.

Volume of a cylinder is area of its base times the height. Area of its base, a circle (since it's a cylinder) is π*r^2 where r is the radius and π is roughly 3.14. I will leave π as the symbol until the end so it's easier to deal with.

So volume = area * height = πr^2*h = π4^2*10 = 160π (I moved pi to the back). It's units are cm^3 since both measurements were in cm. You can feel free to multiply π by 160 now if you like. Anyway, now that we have the volume we multiply it by the density. so 160π cm^3 * 11.34 g/cm^3 = 1814.4π g. If you multiply that by 3.14 for π you get 5697.216 g. And there's the mass. Let em know if there was something you didn't understand.

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if you add 4.21 mL of solution from a buret into a flask that already contained 80.4 mL of solution, what is the total volume of

NNADVOKAT [17]

Answer:

84.6 mL

Explanation:

To get the total volume of soultion we must add the given volumes together.

4.21 mL + 80.4 mL = 84.6 mL

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3 years ago

A sample may contain any or all of the following ions: Hg22+, Ba2+, Mn2+. No precipitate formed when an aqueous solution of NaCl

Bess [88]

Answer:

Only Mn⁺² is present.

Explanation:

When an aqueous solution of NaCl is added, the Cl⁻ species is introduced. In the presence of Cl⁻, Hg₂²⁺ would precipitate as Hg₂Cl₂.

When an aqueous solution of Na₂SO₄ is added, the SO₄⁻² species is introduced. In the presence of SO₄⁻², Ba⁺² would precipitate as BaSO₄.

No precipitate formed when either of these solutions were added, thus <u>the sample does not contain Hg₂⁺² nor Ba⁺²</u>.

Under basic conditions Mn⁺² would precipitate as Mn(OH)₂. A precipitate formed once the solution was made basic, so <u>the sample contains Mn⁺²</u>.

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3 years ago

If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79

Helen [10]

1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass.

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

$200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4$

3) convert moles of CuSO₄ to moles of Cu

$1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu$

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

$1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu$

I did it step-by-step as the explanation but you can do all of this in one step.

$200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu$

4 0

4 years ago

Why blowing at the ping ping ball did not change its shape and size?

goldfiish [28.3K]

Answer:

Because it does not have to

Explanation:

Why because when you blow at it it might move it does not change it's shape because it is a solid

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2 years ago

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Vinil7 [7]

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