8. Extinction of a species could result from
1 answer:
You might be interested in
Answer: Option D is correct.
Explanation: Equation given by de Broglie is:
where, = wavelength of the particle
h = Planck's constant
m = mass of the particle
v = velocity of the particle
In option A, football will have some mass and is moving with a velocity of 25 m/s, hence it will have some wavelength.
In Option B, unladen swallow also have some mass and is moving with a velocity of 38 km/hr, hence it will also have some wavelength.
In Option C, a person has some mass and is running with a velocity of 7 m/hr, hence it will also have some wavelength.
As, the momentum of these particles are large, therefore the wavelength will be of small magnitude and hence, is not observable.
From the above, it is clearly visible that all the options are having some wavelength, so option D is correct.
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
