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djverab [1.8K]
3 years ago
12

Is cutting your hair a physical or chemical change

Chemistry
2 answers:
nalin [4]3 years ago
7 0
Physical because you are physically doing it
Vadim26 [7]3 years ago
7 0
A physical change because you are not changing the chemical composition of your hair. In theory, you could still put that hair back onto your head and it would match the hair still on your head. If it was chemically changed, you would not be able to replace it. Think of wood burning; you can't turn those ashes back to wood, now can you?
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A new grill has a mass of 30.0 kg. you put 1.5 kg of charcoal in the grill. you burn all the charcoal and the grill has a mass o
atroni [7]

The reaction for burning of charcoal or complete combustion is as follows:

C(s)+O_{2}(g)\rightarrow CO_{2}(g)

From the above balanced reaction, 1 mole of carbon releases 1 mole of CO_{2} gas.

Converting mass of charcoal into moles as follows:

n=\frac{m}{M}

Molar mass of pure carbon is 12 g/mol thus,

n=\frac{1.5\times 10^{3} g}{12 g/mol}=125mol

The same moles of CO_{2} is released. Converting these moles into mass as follows:

m=n×M

Molar mass of CO_{2} is 44 g/mol thus,

m=125mol\times 44 g/mol=5.5\times 10^{3}g

Converting mass into kg,

1g=10^{-3}kg

Thus, total mass of gas released is 5.5 kg.

7 0
3 years ago
I need help filling out nitrogen
xxMikexx [17]
For Nitrogen Atom:
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Protons - 7
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Electrons - 7
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For Nitrogen Ion:
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5 0
2 years ago
Why are the oxidation and reduction half-reactions separated in an<br> electrochemical cell?
Mrac [35]

Answer:

The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

7 0
3 years ago
Read 2 more answers
Stone tools enabled people to hunt and grow crops made the food supply more
Nastasia [14]
It makes it more accessible to everybody
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
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