Answer:transpiration
Explanation:
The water eventually is released to the atmosphere as vapor via the plant's stomata — tiny, closeable, pore-like structures on the surfaces of leaves. Overall, this uptake of water at the roots, transport of water through plant tissues, and release of vapor by leaves is known as transpiration.
Answer:- The pressure of ethanol would be 109 mmHg.
Solution:- This problem is based on Clausius clapeyron equation--
Given, 
= 63.5 + 273 = 336.5 K
= 34.9 + 273 = 307.9 K
= 400 mmHg
= ?
= 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/mol.K
Let's plug in the values in the equation and do the calculations.
= 1.30
On taking anti ln to both sides...
= 
= 3.67
= 400/3.67
= 109 mmHg
A.) Cesium is the lowest electronegative element......
Answer:
Kp = 1.39 x 10⁶
Explanation:
The strategy here is to utilize the van´t Hoff relation to calculate Kp at 3 ºC:
ln ( K₂ / K₁ ) = - ΔHºrxn / R x [1/T₂ - 1/T₁]
where K₂ and K₁ are the equilibrium constants at temperatures T₂ and T₁ , ΔHºrxn is the enthalpy change for the reaction and R is the gas constant.
Thus,
Let K₂ equal the equilibrium constant at 3ºC ( 3 + 273 ) K, then:
T₂ = 276 K
T₁ = (25 + 273) K = 298 K ( standard temperature in thermodymics)
ΔHºrxn = -128 kJ/mol x 1000 J /kJ = -1.28 x 10⁵ J / mol
ln ( K₂ / 2.25 x 10⁴ ) = - (-1.28 x 10⁵ J / mol/ 8.314 J/K mol ) x [ 1 / 276 K - 1 / 298 K]
= 4.12
Taking inverse natural log function to both sides of the equation,
K₂ / 2.25 x 10⁴ = e^(4.12)
K₂ = 1.39 x 10⁶
