Answer:
V₂ = 4.34 L
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 3.50 L
Initial pressure = 150 Kpa (150/101.325 = 1.5 atm)
Initial temperature = 330 K
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1.5 atm × 3.50 L × 273 K / 330 K × 1 atm
V₂ = 1433.3 atm .L. K / 330 k.atm
V₂ = 4.34 L
Water is known as a life-giving liquid because every living organism depends on the water for its survival.
<h3>Journal about the topic Water-The Life Giving Liquid</h3>
Water is also called a life-giving liquid because without water life on earth is not possible. All other living organisms need to consume water to survive and grow in size. We need water for many other activities such as cleaning, washing, cooking and irrigation. Water is essential for all living things including humans, animals and plants. Water is called wonder liquid as it can dissolve large number of substances. This property enables water to be a great solvent. Most of the chemical reactions of the living world is carried out in water as a medium so that's why water is known as Life Giving Liquid.
So we can conclude that Water is known as a life-giving liquid because every living organism depends on the water for its survival.
Learn more about water here: brainly.com/question/1313076
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D is the only choice here that is appropriate given neutrons' lack of charge
The answer would be the amino group and the carboxyl group
hope this helpssss
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
