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Dominik [7]
3 years ago
15

Please help I’m torn between A and B. I’ve asked many and people say A, but I asked my science teacher and she said B. And also

I’ve researched and on answers.com also says Water is the better insulator..but idk I’m conflicted! APEX But then again it says
THERMAL INSULATOR...

Chemistry
1 answer:
astraxan [27]3 years ago
4 0
If your science teacher says B, it’s probably because water has a negative and positive end, heat is just a form of energy, as other atoms can’t leave (they’re attracted to the ends) they are being insulated; but notice that ice will melt into gas (where atoms have tons of space) for other atoms to escape. Hence ice and gas aren’t ideal. (Air is a gas here.)
It’s not a 100% but hopefully it helps with some kind of analogy.
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C. Ne

down here is the list of monoatomic elements, just for you :)

4 0
3 years ago
17.8g of aluminum sulfate is produced from how many grams of potassium sulfate?​
german

Answer:

1 mole of ferric contains 2 moles of iron,and 12 moles of oxygen atoms, and three moles of sulphate ions

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.

3 0
3 years ago
What are 3 characteristics of cnidarians?
patriot [66]
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5 0
3 years ago
Write the NET IONIC EQUATION between iron and copper (II) sulfate.
ratelena [41]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of iron and copper (II) sulfate is given as:

Fe(s)+CuSO_4(aq.)\rightarrow FeSO_4(aq.)+Cu(s)

Ionic form of the above equation follows:

Fe(s)+Cu^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow Fe^{2+}(aq.)+SO_4^{2-}(aq.)+Cu(s)

As, sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Fe(s)+Cu^{2+}(aq.)\rightarrow Fe^{2+}(aq.)+Cu(s)

Hence, the net ionic equation is written above.

7 0
3 years ago
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