The partial pressure of oxygen in a sample of air increases if the temperature is increased.
Answer: Option 1
<u>Explanation:
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According to Guy-Lussac's law, at constant volume, pressure exhibited by the gas molecules will be directly proportional to the temperature of the gas molecules. It is also known that pressure of mixture of gas molecules is the sum of partial pressure of each gas molecule in the mixture.
If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.
So, it can also be concluded that temperature increases the oxygen's partial pressure in air increases.
Answer:
Condensation
Explanation:
An exothermic process is one in which heat is lost. Condensation is a change of state from gas to liquid. Thus is loses heat.
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Answer:

Explanation:
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In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

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Answer:
They all wear their skeletons on the outside! This is called an exoskeleton and all creatures that have their structure on the outside are included in the phylum of Arthropods.
Explanation:
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ