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3 years ago

What happened to the number of moles in a sample that originally occupied 500mL with 2.50 moles and then occupied 750mL?

2 answers:

7 0

Either you have added an extra half of the # of moles or you changed the condition (lower the pressure or increase the temp.)

ANEK [815]3 years ago

3 0

Answer: The number of moles that initially occupied a 250ml container remains the same when it occupied 750ml container but they are freer because the 750ml container is bigger and has more space. So the gases now occupied more space and their movement is less curtailed unlike when they were in 250ml container

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Unit 1- Unit Assessment (Matter & Density)

konstantin123 [22]

Answer:

<h3>The answer is 11 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

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

From the question

mass = 3025 g

volume = 275 mL

We have

$density = \frac{3025}{275} \\$

We have the final answer as

Hope this helps you

7 0

2 years ago

I’ll give the brainliest to whoever gets it right!!!

kramer

Answer:

itll slowly decay

Explanation:

i hope this helps

8 0

2 years ago

Use Boyle's Law to explain what would happen to the volume of a helium- filled balloon if it was carried underwater by a diver t

oksano4ka [1.4K]

The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.

4 0

2 years ago

Lead has a density of 11.29 g/mL. If you had 186 g of lead, what would its volume be?

Varvara68 [4.7K]

I think you just do 11.29 multiplied by 186 since to find density you divide mass by volume. So 186 divided by x is 11.29. So in conclusion the volume would be 2,099.94mL

8 0

2 years ago

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?

NARA [144]

Answer:

ΔH = $q_{p}$

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

$q_{com}$ = $q_{p}$

where

$q_{p}$ = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

$q_{com}$ = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is $q_{p}$

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = $q_{p}$ + 0

==> ΔH = $q_{p}$

5 0

3 years ago