Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
The proximate reason for the uptake by plants of nutrients like ammonium nitrate is homeostatis
<h3>
What is homeostasis?</h3>
Homeostasis refers to an organism's ability to regulate various physiological processes to keep internal states steady and balanced.
<h3>How does plant perform homeostatis?</h3>
As water leaves the plant tissues into the atmosphere, it takes energy with it in the form of heat.
Much like when we sweat, this allows the plant to cool and maintain homeostasis.
Similarly, the same process allows plants to absorb nutrient like ammonium nitrate.
Thus, the proximate reason for the uptake by plants of nutrients like ammonium nitrate is homeostatis.
Learn more about homeostatis here: brainly.com/question/1046675
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Answer:
Explanation:
Using freezing point depression formula,
ΔTemp.f = Kf * b * i
Where,
ΔTemp.f = temp.f(pure solvent) - temp.f(solution)
b = molality
i = van't Hoff factor
Kf = cryoscopic constant
= 1.86°C/m for water
= (0 - (-5.58))/1.86
= 3.00 mol/kg
Assume 1 kg of water(solvent)
= (3.00 x 1)
= 3.00 mol.
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
