Answer: Transpiration---A
Explanation: Transpiration is the process in the water cycle whereby plant loose(excess) water by evaporation through the stomata of their leaves since not all water absorbed by the root is actually used for growth in plants.In order to allow the intake of carbon-dioxide, water must exit the leaves through transpiration which then provides the plant with cooling, rigidity and maintaining the overall water balance of the plant.
A. a wide variety of hunting techniques
Explanation:
To help the fisher population to survive in a changing environment, they have to develop a wide variety of hunting techniques.
The new fisher population have to develop a wide range of strategies to satisfy its carnivorous instincts.
- Most predators are carnivores with a wide range of hunting skills
- They must be fast, subtle, good sighted and well adapted body part to hunt of preys.
- As they transcends different environments, they must have good and diverse hunting skills to give them a competitive advantage.
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C is the right answer as endothermic reaction takes in the heat. That is why cooling effect is observed whenever such reaction takes place.
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Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that occupies.
- T is its temperature.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
Then, in this case:
- P= 1 atm
- V= 44.1 L
- n= ?
- R= 0.082

- T= 0°C =273 K
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082
× 273 K
Solving:

n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
Answer:
.02232 mole
Explanation:
At STP each 22.4 liters is a mole
.5 liters / 22.4 l/mole = .02232 mole