Chemistry, 25.08.2020 07:01 candlewax2001

Find the density of a block with a length of 8.0 cm, a width of 4.0 cm, a height 2.0 cm, and a mass of 32 g. Would this block
float or sink in water?

Answers: 2

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Answer from: Galaxy6267

Density = 0.5 g/cm³

Explanation:

Density of a substance can be found by using the formula

$Density = \frac{mass}{volume}$

From the question

mass = 32g

To find the density we must first find the volume of the block

From the question the block is a cuboid

Volume of cuboid = length × width × height

length = 8 cm

width = 4 cm

height = 2 cm

Volume of block = 8 × 4 × 2 = 64 cm³

Substitute the values into the above formula and solve for the density

That's....

$Density = \frac{32}{64} \\ = \frac{1}{2}$

We have the final answer as

Density = 0.5 g/cm³

The block will float because it's density is less than the density of water which is

1 g/cm³

Hope this helps you

Answer from: noathequeen

d= 0.5 g/ cm³

The block will float in water.

Explanation:

The density of an object can be found using the following formula:

$d=\frac{m}{v}$

where is the mass and is the volume. We know the block has a mass of 32 grams. We don't know the volume, therefore we must calculate it.

Let's assume the block is a cube. The volume of a cube can be found using the following formula.

v=l*w*h

The length is 8 cm, the width is 4 cm and the height is 2 cm.

$l= 8 cm \\w= 4cm \\h= 2cm$

v= 8 cm * 4 cm * 2cm

v=32 cm^2 * 2cm

v= 64 cm^3

Now we know the volume and can substitute it into the formula, along with the mass.

$d=\frac{m}{v}$

$m= 32 g\\v= 64 cm^3$

$d=\frac{32 g}{64 cm^3}$

Divide 32 grams by 64 cubic centimeters.

d= 0.5 g/cm^3

The density of the block is 0.5 grams per cubic centimeter. The density of water is 1.0 g/cm³. Any object with a density that is less than 1 will float in water. 0.5 is less than 1, so this object will float in water.

Answer from: Quest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

Answer from: Quest

i would say b

explanation:

but i think i would need more information than that

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Find the density of a block with a length of 8.0 cm, a width of 4.0 cm, a height 2.0 cm, and a mass...

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