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e-lub [12.9K]
3 years ago
9

Please can u help me .

Chemistry
2 answers:
Marysya12 [62]3 years ago
7 0

Answer:

4 is D 5 is A

Explanation:

sdas [7]3 years ago
4 0

QUESTIONS !

  • 1. Why might pesticides sprayed onto the fields reduce a bean crop?

A Pesticides damage plant leaves.

B Pesticides damage plant roots.

C Pesticides kill insects that feed on bean plants.

D Pesticides kill insects that pollinate bean plants.

  • 2. What may cause the rapid growth of plants in a lake?

A excess fertilizers

B excess herbicides

C low water pH

D low water temperature

ANSWER !

  • 1. (D) Pesticides kill insects that pollinate bean plants.
  • 2. (B) Excess Herbicides

EXPLANATION !

  • 1. When a field is sprayed with a pesticide, not all of the insects will die, and the ones that remain will produce offspring that will also be tolerant to the spray. This forces farmers to use more and different kinds of pesticides, so they can kill all of the insects that might destroy their crop.
  • 2. During periods of heavy rainfall, increased runoff brings more nutrients into a lake. These additional nutrients may be released from flooded soils or carried into the lake on silt or clay particles. Excess nutrients often boost plant growth, particularly algae or free-floating aquatic plants.                       <em>PLEASE MARK AS BRAINLIEST !!!!</em>
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7 0
2 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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4 years ago
A simple way to look at kinetic energy is
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