Things an animal does to survive are called behavioral adaptations. The correct option is c.
<h3>What is behavioral adaptation?</h3>
The adaptation of behavioral and physical characters that help to survive them.
For example, hibernating during winter, migration, etc. There are two types of behavioral adaptation, learned and instinctive.
Thus, the correct options are c: behavioral adaptations.
Learn more about behavioral adaptation
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Answer:
The volume is 1.2L
Explanation:
Initial volume (V1) = 700mL = 0.7L
Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K
Initial pressure = 106.6kPa = 106600Pa
Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K
Final pressure (P2) = 66.6kPa = 66600Pa
Final volume (V2) = ?
To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.
(P1 × V1) / T1 = (P2 × V2) / T2
solve for V2 by making it the subject of formula,
P1 × V1 × T2 = P2 × V2 × T1
V2 = (P1 × V1 × T2) / (P2 × T1)
V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)
V2 = 22397193 / 18657990
V2 = 1.2L
The final volume of the gas is 1.2L
Answer:
Heating this gas to 55 °C will raise its volume to 6.87 liters.
Assumption: this gas is ideal.
Explanation:
By Charles's Law, under constant pressure the volume 
of an ideal gas is proportional to its absolute temperature 
(the one in degrees Kelvins.)
Alternatively, consider the ideal gas law:
.
is the number of moles of particles in this gas. should be constant as long as the container does not leak.
is the ideal gas constant. 
is the pressure on the gas. The question states that the pressure on this gas is constant.
Therefore the volume of the gas is proportional to its absolute temperature.
Either way,
.
.
For the gas in this question:
- Initial volume:
.
Convert the two temperatures to degrees Kelvins:
- Initial temperature:
. - Final temperature:
.
Apply Charles's Law:
.
Answer:
0.01M = [H⁺]; 1x10⁻¹²M = [OH⁻]; Ratio is: 1x10¹⁰
Explanation:
pH is defined as -log [H⁺]
For a pH of 2 we can solve [H⁺] as follows:
pH = -log [H⁺]
2 = -log [H⁺]
10^-2 = [H⁺]
<h3>0.01M = [H⁺]</h3>
Using Keq of water:
Keq = 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 0.01M = [OH⁻]
<h3>1x10⁻¹²M = [OH⁻]</h3><h3 />
The ratio is:
[H⁺] / [OH⁻] = 0.01 / 1x10⁻¹² =
<h3>1x10¹⁰</h3>
