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3 years ago

Which phase change requires more heat energy, Heat of fusion or Heat of

Chemistry

2 answers:

Afina-wow [57]3 years ago

5 0

Answer:

heat of fusion

Explanation:

heat of fusion is melting which requires more heat energy than heat of vaporization

OverLord2011 [107]3 years ago

3 0

Answer:

Evaporation involves a liquid becoming a gas and sublimation is the change of a solid directly to a gas. Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing).

Explanation:

You might be interested in

What are the half-reactions for a galvanic cell with aluminum and gold electrodes?

pashok25 [27]

Answer:B is right

Explanation: Al loses electrons and is oxidised, Au gains electron. And is reduced

4 0

3 years ago

Cân bằng phương trình Mg+HCl->MgCl2+H2 bằng phương pháp thăng bằng

nignag [31]

quá trình oxi hóa: Mg0 -> Mg2+ + 2e-

quá trình khử: 2H+ + 2e- -> H2 (cần 2 H+ để tạo 1 H2)

----> chỉ cần thêm 2 vào trước HCl (vì H+ là do HCl cung cấp)

----> Mg + 2HCl -> MgCl2 + H2

yay :)

8 0

2 years ago

What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?

____ [38]

Cell

2.115

Cathode

Anode

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

(

)

−

→

(

)

−

2.372

(

)

−

→

(

)

−

0.257

The standard reduction potential

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

1.00

kPa

) is defined as

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

will experience oxidation and act the anode.

(

)

(

)

Therefore in this galvanic cell, the

half-cell will experience reduction and act as the cathode and the

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

cell

(

Cathode

)

−

(

Anode

)

cell

−

0.257

−

(

−

2.372

)

cell

2.115

Indicating that connecting the two cells will generate a potential difference of

2.115

across the two cells.

5 0

2 years ago

Read 2 more answers

[Then mother noticed that two children who are watching the game are sick and coughing.She told her son to keep distance from th

Andrew [12]

Answer:

in illustration 1

the mother told her son not to play with the boy having cough because it is a transferable disease which could cause him also to get sick and would get cough.

while in

illustration 2

the mother let the boy play with the boy having asthma because it is not a transferable disease which wold not affect him or any one else.

Explanation:

this is the reason

7 0

2 years ago

Read 2 more answers

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

8 0

3 years ago