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sesenic [268]
2 years ago
6

You have a substance with a length of 2 cm, width of 3 cm, and a height of 4 cm. Its density is known to be .5 g/cm3. What is th

e mass of the sample you have?
Chemistry
1 answer:
otez555 [7]2 years ago
5 0

Explanation:

They are related by the the density triangle.

Explanation:

They are related by the the density triangle.

mcdn1.teacherspayteachers.com

d =

m

V

m = d×V

V =

m

d

DENSITY

Density is defined as mass per unit volume.

d =

m

V

Example:

A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?

Step 1: Calculate the volume

V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³

Step 2: Calculate the density

d =

m

V

=

433

g

200

c

m

³

= 2.16 g/cm³

MASS

d =

m

V

We can rearrange this to get the expression for the mass.

m = d×V

Example:

If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?

m = d×V = 500 mL ×

1.11

g

1

m

L

= 555 g

VOLUME

d =

m

V

We can rearrange this to get the expression for the volume.

V =

m

d

Example:

What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.

Step 1: Convert kilograms to grams.

14.83 kg ×

1000

g

1

k

g

= 14 830 g

Step 2: Calculate the volume.

V =

m

d

= 14 830 g ×

1

c

m

³

19.32

g

= 767.6 cm³

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During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that
bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

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svetlana [45]

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