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DerKrebs [107]
3 years ago
11

Which ion has a smaller size,1)N---,O--,F-and Ne 2) Cl-,Br- and I ?Why Help me to solve it!

Chemistry
2 answers:
Vlada [557]3 years ago
8 0

Explanation:

  1. Ne have smaller size because Ne have no negative charge compare to N, O and F which negative charge depends on size of an atom, Therefore more the negative charge , more will be the size of atom.
  2. Cl- have smaller size because as we go down the group, Size of the atom increase as Cl<Br<I irrespective to the atomic charge
sineoko [7]3 years ago
3 0

Answer:

nitrogen because it has less number of protons and electrons, thus a samller atomic mass

Explanation:

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Which element is more reactive in water?<br><br> -Na<br><br> -Mg
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The density of water at 4 degrees Celsius is 1.00 g/cm cubed at unknown object has density of 7.9 g/cm cubed would you expect it
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7 0
3 years ago
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
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