1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
notsponge [240]
2 years ago
7

Three of the adaptations would be very important to help a temperate forest plant disperse its population and spread out. One ad

aptation does not aid in population dispersal. That adapatation is
Chemistry
2 answers:
yan [13]2 years ago
8 0

Answer:

The correct answer is "the presence of secondary xylem as wood tissue".

Explanation:

The missing options of this question are:

A. runner roots that produce new plants by vegetative propagation

B. the use of large sugary fruits to attract animals

C. the presence of secondary xylem as wood tissue

D. the use of pollinating insects, such as bees

The correct answer is option C. "the presence of secondary xylem as wood tissue".

Temperate forest plants have a few adaptations that help them to disperse its population, however, the presence of secondary xylem as wood tissue is not one of them. The adaptation of developing a secondary xylem as wood tissue serves the plant as a defense system. Numerous studies report that trees with secondary xylem as wood tissue have less propensity of developing fungal and bacterial infections.

Likurg_2 [28]2 years ago
3 0

Answer:

C, <u>The presence of secondary xylem as wood tissue.</u>

Explanation:

The presence of secondary xylem as wood tissue. While this strengthens a plant and allows it to grow taller, it does not directly help a plant to spread.

You might be interested in
Addie mixes together a solution of 10 grams of
Gnesinka [82]

Answer: 150 grams

Explanation: m = V × ρ

= 15 milliliter × 10 gram/cubic centimeter

= 15 cubic centimeter × 10 gram/cubic centimeter

= 150 gram

5 0
3 years ago
Amount of liquid measured in L or Lm
devlian [24]

The amount of liquid is measured in L. As the unit Lm is only used when measuring the mass liquid fraction.


5 0
3 years ago
Quick electron emissions are called
Montano1993 [528]
Defined as a phenomenon of liberation of electron from the surface that is stimulated by temperature elevation, radiation, or by strong electric field.
3 0
3 years ago
The distance from the sun to the earth would be which phrase best completes the sentence
Neporo4naja [7]

Answer:

The distance from Earth to the sun is called an astronomical unit, or AU, which is used to measure distances throughout the solar system.

6 0
3 years ago
Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0
3 years ago
Other questions:
  • Which of the following describes a condition in which an individual would not hear an echo?
    10·2 answers
  • Which of the following is a precipitation reaction?
    6·1 answer
  • Which subatomic particle makes the atoms of each elements different?
    5·1 answer
  • Suppose that you have 125 mL of a buffer that is 0.140 M in both hydrofluoric acid ( HF ) and its conjugate base ( F − ) . Calcu
    5·1 answer
  • I need help with number two too?
    11·1 answer
  • Emission results in the release of energy without any apparent change in mass or nuclear charge.
    6·1 answer
  • Which is true of an element?
    6·2 answers
  • Which Earth motion describes the wobbling of the Earth around its precessional axis and
    12·1 answer
  • Scientist repeats an experiment and gets a<br> different result. What should the scientist do next
    10·1 answer
  • 78.5 g of Carbon dioxide gas fills a container that is 1.2 L. What is the
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!