Answer:
Excess Reagent = oxygen
Explanation:
Limiting reagent: The substance that is totally consumed when the reaction is completed.
Excess reagent: The substance left after the limiting reagent is consumed completely
The balanced chemical equation for formation of water is as follow:
This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.
Hence the ratio in which hydrogen and oxygen gas reacts is 2:1
Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.
or
Here 5 mole of oxygen is reacting but only 2 mole is required .
Oxygen is in excess.
In the so called rain shadow effect we have interaction between all of the four major Earth spheres. When we have a coastal region where there's a high mountain range, the part of the mountain that is facing the sea will differ a lot from the part of the mountain that is on the other side. The water from the sea evaporates. The water vapor makes the air wet. The warm and wet air masses from the sea will come to the coastline, once they reach the mountain they will start to accumulate as they can not pass through it. As they accumulate rainfall appears. The rainfall contributes to a lush vegetation on this side of the mountain (windward side). The rain shadow effect appears on the leeward side of the mountain, and it mostly gets dry, strong, downward winds. These conditions result in drier climate, much less vegetation, and much increased erosion. Thus we can easily see that we have in this case interaction between the hydrosphere (the sea and the rainfall), the geosphere (the ground, soil, rocks), biosphere (the vegetation), and atmosphere (the winds, the clouds).
Answer:
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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
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