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NeX [460]
3 years ago
8

A plant release it’s seed , but soon there after, the area seeds are in experiences a yearlong drought. How would the seeds most

likely respond to this change in the environment?
Chemistry
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.

Explanation:

Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.

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3 years ago
The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
Brilliant_brown [7]

Answer :  The rate of reaction is,

Rate=4.77\times 10^{-19}M/s

The appearance of NO_3 is, 4.77\times 10^{-19}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)

The rate law expression will be:

Rate=k[NO_2][O_3]

Given:

Rate constant = k=1.69\times 10^{-4}M^{-1}s^{-1}

[NO_2] = 1.77\times 10^{-8}M

[O_3] = 1.59\times 10^{-7}M

Rate=k[NO_2][O_3]

Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})

Rate=4.77\times 10^{-19}M/s

The expression for rate of appearance of NO_3 :

\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}

As, \text{Rate of reaction}=4.77\times 10^{-19}M/s

So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

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