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Vanyuwa [196]
3 years ago
10

As sample of gas occupies 2.00L with 5.00 moles present. What would happen to the volume if the number of moles is increased to

10.0 miles?
Chemistry
1 answer:
maw [93]3 years ago
8 0
The volume would increase and it will become 4.00 L
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Which statement BEST explains why liquids and gases can flow but solids cannot?
storchak [24]

Answer:

B

Explanation:

A is not the answer.  Although the statement is accurate in regards to gases, it does not explain why liquids and gases can flow.

B is the answer.  Solids are in fixed structures.  When you apply heat or pressure, these structures are broken apart and allowed to move freely.

C is not the answer.  This is inaccurate.  Changing the phase of a substance does not change the size of the particle.

D is not the answer.  The opposite of this statement is true.  The attractive forces between particles in a solid allow the substance to hold its structure.  When you apply heat or pressure, the attractive forces are overpowered and the structure is broken.

7 0
3 years ago
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1. The volume of a given mass of gas is 720 ml at 15°C. Assuming
Anni [7]

please mark my answer brainliest...

condition...pressure remains constant.

  1. for 720 ml temp is 15°...so for
  2. 960ml temp will be 15/720×960=20°...(.answer for 1st part...)
  3. for 900cmcube temp is 270°C...so for
  4. 300cmcube temp will be 270/900×300=90°....(answer for 2nd part)...
  5. I hope it helps the dear students...and if it is then let me know through ur comments...and please mark my answer as brainliest...plz...
4 0
3 years ago
The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.
atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M  

Explanation:

<u>The photodissociation of ozone by UV light is given by:</u>

O₃ + hν → O₂ + O (1)

<u>The first-order reaction of the equation (1) is:</u>

rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t} (2)

<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>    

<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>

[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt} (3)

<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>

We can calculate the initial ozone concentration using equation (3):  

[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!                    

3 0
4 years ago
The body of knowledge that deals with Earth and its place in the universe is called _____________?
lakkis [162]

Answer:

Cosmology is a branch of astronomy that involves the origin and evolution of the universe, from the Big Bang to today and on into the future. According to NASA, the definition of cosmology is "the scientific study of the large scale properties of the universe as a whole."

7 0
3 years ago
Which is a reason to use rate laws?
vekshin1

Answer:

B

Explanation:

to find the new rate when the concentration of reactants changes

5 0
3 years ago
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