Answer:
condensation
Explanation:
Frost forms when an outside surface cools past the dew point. The dew point is the point where the air gets so cold, the water vapor in the atmosphere turns into liquid. This liquid freezes. If it gets cold enough, little bits of ice, or frost, form
Answer:
1750L
Explanation:
Given
Initial Temperature = 25°C
Initial Pressure = 175 atm
Initial Volume = 10.0L
Final Temperature = 25°C
Final Pressure = 1 atm
Final Volume = ?
This question is an illustration of ideal gas law.
From the given parameters, the initial temperature and final temperature are the same; this implies that the system has a constant temperature.
As such, we'll make use of Boyle's Law to solve this;
Boyle's Law States that:
P₁V₁ = P₂V₂
Where P₁ and P₂ represent Initial and Final Pressure, respectively
While V₁ and V₂ represent Initial and final volume
The equation becomes
175 atm * 10L = 1 atm * V₂
1750 atm L = 1 atm * V₂
1750 L = V₂
Hence, the final volume that can be stored is 1750L
Hope my answer helped you
Answer:
<u>It increases by a factor of four</u>
Explanation:
Boyle's Law : At constant temperature , the volume of fixed mass of a gas is inversely proportional to its pressure.
pV = K.......(1)
pV = constant
Charles law : The volume of the gas is directly proportional to temperature at constant pressure.
V = KT
or V/T = K = constant ....(2)
Applying equation (1) and (2)
According to question ,
T2 = 4 (T1)
V2 = V1
Put the value of T2 and V2 , The P2 can be calculated,
V1 and V1 cancel each other
T1 and T1 cancel each other
We get,
or
P2 = 4 P1
So pressure increased by the factor of four
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
brainly.com/question/17088180
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