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STatiana [176]
2 years ago
12

A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

L. What was the initial volume of the balloon?
Chemistry
1 answer:
aivan3 [116]2 years ago
4 0

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = n_1=9.79 mol

Initial volume of freon gas in ballon = V_1=?

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol

Final volume of freon gas in ballon = V_2=21.8 L

Using Avogadro's law:

\frac{V_1}{n_1}=\frac{V_2}{n_2} ( at constant pressure and temperature)

V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L

16.06 L was the initial volume of the balloon.

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gizmo_the_mogwai [7]

Answer:

It is because the object that is charged attracts the uncharged because it is giving electrons and passing it on to the object. As shown in the image the balloon is attracting the paper that is not charged. Actually telling,the charged object will cause the uncharged object to become charged, with a positive charge on one side and a negative charge on the other side. This process is called induction.

Explanation: Pls Mark Brainliest

4 0
2 years ago
What is the correct chemical equation for the reaction between methane and oxygen to produce carbon dioxide and water?
kodGreya [7K]

Answer:

The answer to your question is:   CH₄     +    3/2 O₂     ⇒     CO₂    +   2 H₂O

Explanation:

Methane = CH₄

Oxygen = O

Carbon dioxide = CO₂

Water = H₂O

                      CH₄     +    3/2 O₂     ⇒     CO₂    +   2 H₂O

        This is the balanced equation

                 

3 0
3 years ago
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr
lys-0071 [83]

Answer:

a. No change.    

b. The equilibrium will shift to the right.

c. No change

d. No change

e.  The equilibrium will shift to the left

f.  The equilibrium will shift to the right      

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining   Keq  again.

The equilibrium constant  for  A(s)⇌B(g)+C(g)  

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use  Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a.  double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´     ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp  ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to  double the pressures of A and B. Therefore there is no change.

b.  double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change  reduced the pressures by a half :

Q =  pB´x pC´  = (  1/2 pB ) x ( 1/2 pC )  =  1/4 pB x pC  = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the  concentration of B and halve the concentration of C

Doubling the concentrantion doubles  the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q =  pB´x pC´  = ( 2 pB ) x ( 1/2 pC ) = K

e.  double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q =  pB´x pC´   = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f.  double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q =  pB´x pC´   = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16  Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

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3 years ago
What is the main difference between electron configuration and orbital notation?
8_murik_8 [283]

Answer:

Orbital Notation is more specific on where exactly the electron is placed.

Explanation:

When writing an electron configuration for an atom, rather than writing out the occupation of each and every orbital specifically, you instead lump all the core electrons together and designate it with a symbol of the corresponding noble gas on the Periodic Table.

the arrangement of electrons in the orbitals of an atom or molecule

While Orbital Notation is a visual transformation of the electron configuration. It shows you where each specific electron is placed and what its "spin" is.

Glad I could help!

6 0
3 years ago
PLEASE HELP!!!! <br>it would mean alot ! ​
statuscvo [17]

Answer:

i looked it up its slinky waves

Explanation:

6 0
3 years ago
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