Question

answer please fast it's important​

Answer 1

Answer:

The leaves of certain plants exhibit droplets of water along their margins in the morning. This particularly happens in plants growing in warm humid conditions. A humid environment hampers transpiration while the roots continue to absorb water from the soil. This builds up a big hydrostatic pressure within the plant and “forces out” the excess water directly from the tips of veins in the leaf. Special pore-bearing structures called hydathodes are present on the margins of the leaf to allow this exudation.

Answer 2

Answer:Answer

Solution:-

OF  

2

​

 

(g)

​

+H  

2

​

O  

(g)

​

⟶O  

2

​

 

(g)

​

+2HF  

(g)

​

 

ΔH  

R

​

=∑ΔH  

f

​

 

(product)

​

−∑ΔH  

f

​

 

(reactant)

​

 

∴ΔH  

R

​

=(2×(−268.6))−(23+(−241.8))

⇒ΔH  

R

​

=−756kJ/mol

Hence the standard enthalpy change will be −756kJ/mol

Now, as we know that,

ΔH=ΔU+Δn  

g

​

RT

⇒ΔU=ΔH−Δn  

g

​

RT

Now from the given reaction,

Δn  

g

​

=n  

P

​

−n  

R

​

=(2+1)−(1+1)=1

T=300K(Given)

R=8.314×10  

−3

J/mol−K

∴ΔU=(−756)−(1×8.314×10  

−3

×300)

⇒ΔU=−756−2.494=−753.506kJ/mol

Hence the standard internal energy change will be −753.506kJ/mol.

Explanation: