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V125BC [204]
3 years ago
6

A sample of krypton occupies 15.0 L at a pressure of 2.1 atm. Use Boyle's Law to find the pressure of the krypton when the volum

e is decreased to 4L.
Chemistry
1 answer:
denis23 [38]3 years ago
4 0

Answer:

7.5 atm

Explanation:

Given that,

Initial volume, V₁ =  15 L

Pressure, P₁ = 2.1 atm

Final volume, V₂ = 4L

We need to find the final pressure. The mathematical relation between volume and pressure is given by :

\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}\\\\P_2=\dfrac{V_1P_1}{V_2}\\\\P_2=\dfrac{15\times 2}{4}\\\\P_2=7.5\ atm

So, the final pressure is equal to 7.5 atm.

You might be interested in
What mass of ag2co3 could you produce from 12.7 g agno3 assuming that it is the limiting reagent?
zhenek [66]

10.3 g of  Ag_{2} CO_{3} will produce from 12.7 g AgNO_{3}.

In simple terms, a limiting reagent is a reactant that is completely used up in the reaction. 

It is also referred to as a limiting reactant or limiting agent.

Now, according to the question,

Since it is a limiting reagent, it will react fully. 

The mass of AgCO_{3} that will be produced will be:-

Equivalent mole of AgNO_{3} = Equivalent weight of Ag_{2} CO_{3} = 12.7/169.97 = x*2/274x = 10.3 g. 

Hence, 10.3 g of Ag_{2} CO_{3} will be produced.

It is used to restrict the reaction.

It tells you the estimated amount of compound to be used.

It brings quantitative understanding to chemical reactions.

More information on limiting reagents can be found here :

brainly.com/question/11848702

#SPJ4

7 0
2 years ago
The final volume of the solution is 284 mL. What is the concentration of CuSO4 in the final solution, in mol/L?
DochEvi [55]

Answer:

0.0252mol/L

Explanation:

The following data were obtained obtained from the question:

Volume of solution = 284mL = 284/1000 = 0.284L

Mole of CuSO4 = 7.157 × 10^-3 mol

Molarity =?

Molarity = mole/Volume

Molarity = 7.157x10^-3 /0.284

Molarity = 0.0252mol/L

The concentration of the solution is 0.0252mol/L

7 0
3 years ago
galvanic cell is powered by the following redox reaction: (g) (aq) (l)(aq) (s) (aq) Answer the following questions about this ce
kotykmax [81]

Answer:

Explanation:

1. the 1/2 reaction that occurs at the cathode

3Cl2(g) +6e^- -------------> 6Cl^- (aq)

2 the 1/2 reaction that occurs at the anode

2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-      

2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-

E0 = -0.59v

3Cl2(g) +6e^- -------------> 6Cl^- (aq)                                                  

E0 = 1.39v

3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)  

E0cell = 0.80v

5 0
3 years ago
Mercury has a density of 13.6 grams/mL, what is this density in cg/L?
Firdavs [7]
It would be 4.6cgL not sure tho because I didint do that good in this
4 0
3 years ago
How many molecules are in 2.38g of SO2
Murrr4er [49]

2.63 x 10^22 molecules of SO2.

To find this, start with what you know.

2.38g of SO2.

You need to first convert this into Moles since you cannot directly convert grams into molecules. In order to convert grams to moles, you need to find the molecular mass of SO2 - 64.066.

This is because Sulfur has the mass of 32.066 and Oxygen has a mass of 16, but since there are two Oxygen atoms, it's going to be 32.

Your equation should currently appear as so:

2.38g of SO2 = 1 Mole of SO2 / 64.066

Now, you need to convert this to molecules.

Whenever you are searching for molecules without a given amount, you always use Avogadro's number: 6.02 x 10^23

Now, your equation should appear as so:

2.38g SO2 = 1 Mole of SO2 / 64.066 next to a new fraction which is 6.02 x 10^23 / 1 Mole of SO2

Now, multiply across (2.38 x 1 x 6.02 x 10^23). When using Avogadro's number, don't forget to place parenthesis around it.

Then, divide that number by the bottom: 64.066.

Thus, your final answer is that there is 2.63 x 10^22 molecules of SO2.

Don't forget your units!

Hope this helps!

5 0
3 years ago
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