What properties of water would help water travel up the leaf?

1 answer:

8 0

The high surface tension of water pulls the concavity outwards, generating enough force to lift water from the roots to the leaves of plants. Thus, Cohesion, along with adhesion and surface tension creates a capillary action that keeps water molecules interacting and moving through the plants out to the leaf cells.

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Does anyone know this ?Please I need help

Marrrta [24]

Explanation:

Entropy refers to the degree of disorderliness of system. Basically it refers to how random a system is.

Gong through the options.

A. The phase change is from solid to liquid. Liquid has a greater degree of disorderliness compared to solid hence entropy is increasing.

B. The phase change is from solid to gas. Gas particles has a greater degree of disorderliness compared to solid hence entropy is increasing.

C. The phase change is from aqueous to liquid. Aqueous solutions has a greater degree of disorderliness compared to liquid hence entropy is decreasing.

D. The reactant side has; solid + gas. In the product side, there is only gas. The product side has a greater degree of disorderliness hence the entropy is increasing.

E. The entropy of the product is greater than the entropy of the reactants hence entropy is increasing.

F. The phase change is from aqueous to solid. Aqueous solutions has a greater degree of disorderliness compared to solid hence entropy is decreasing.

8 0

2 years ago

Ice melting in a glass of iced tea is an example of a(n) __________ process (not necessarily a chemical reaction).

adell [148]

Its a Physical change

8 0

2 years ago

Two billiard balls having masses of 0.2 kg and 0.15kg approach each other. The first ball having a velocity of 2m/s hit the seco

Gekata [30.6K]

Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

Explanation:

Given: $m_{1}$ = 0.2 kg, $m_{2}$ = 0.15 kg

$v_{1}$ = 2 m/s, $v_{2}$ = 0 m/s, $v'_{1}$ = ?, $v'_{2}$ = 1.5 m/s

Formula used is as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}$

where,

v = velocity before collision

v' = velocity after collision

Substitute the values into above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s$