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3 years ago

What is the density of motor oil in g/mL if 1.25 quart (qt) weighs 2.25 pounds (lb)?

Chemistry

1 answer:

Doss [256]3 years ago

5 0

Answer:

0.86 g/mL

Explanation:

In order to solve the problem, you have to know the formula for "Density." Density refers to "mass per unit volume." It's formula is:

Density (ρ) = $\frac{mass}{Volume}$

Next, we have to convert the units of measurement in accordance to what is being asked in the problem. It is asking for "g/mL," which means we have to convert 1.25 quarts (qt.) to mL and 2.25 pounds (lbs.) to grams.

Let's solve.

1 quart = 946.35 mL

1.25 qt. x $\frac{946.35 mL}{1 qt.}$ = 1,182.94 mL

1 pound = 453.592 grams

2.25 pounds x $\frac{453.592 grams}{1 pound}$ = 1,020.58 grams

Now that we converted them to their appropriate units of measurement, we can now solve for density.

ρ = $\frac{mass}{Volume}$

ρ = $\frac{1,020.58 grams}{1,182.94 mL}$

ρ = 0.86 g/mL

The density of the motor oil is 0.86 g/mL

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Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

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Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

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E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

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Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

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