Global warming is warming to the earth people can stop littering to help it
Answer:
The freshwater sources that are generally in continuous motion and follow a defined path are called streams and rivers.
If I were to improve the lab then I will make the following changes:
- The experiment aimed to observe and model the effects of rivers on erosion. So, I can make a virtual model of the river and can compare the velocity, gradients and volume of rivers.
- Comparison between the low and high factors listed can help in computing the effect of the powerful river on erosion.
- The high velocity. gradient and volume of the river will cause more erosion as it exerts more force.
- The low volume, gradient and velocity river will affect in a less manner on erosion.
Explanation:
thats all i know ( correct me if im wrong please)
Hemoglobin has a much greater affinity for carbon monoxide than oxygen. In a hyperbaric chamber (containing high levels of oxygen) can treat carbon monoxide poisoning, by displacing carbon monoxide from Hemoglobin competitively.
Hemoglobin has a much greater affinity for carbon monoxide than oxygen. This is because, a coordinate bond is formed with Carbon monoxide and Haem structure of the hemoglobin.
Carbon monoxide with Hemoglobin is called as Carboxy haemoglobin.
Presence of oxygen displaces the Carbon monoxide with Hemoglobin that is formed due to poisoning.
Hyperbaric chamber is a chamber which contains pure oxygen in a chamber. The atmospheric pressure is kept about three to four times than the normal, such that the replacement of Carbon monoxide from Haem can occur as fast as possible since this reduces the half life of the Carboxy haemoglobin.
It is advisable not to treat Carbon monoxide poisoning yourself.
Hyperbaric oxygen is used to treat the following conditions as well:
- Infections
- Wounds
- Air bubble is blood
Learn more about Carbon Monoxide here, brainly.com/question/11313918
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Answer:
37.14 %
Explanation:
Using the equation, mass, M = D1 * V1
= D2 * V2
Where,
D1 = density of the liquid Nitrogen
D2 = density of gaseous Nitrogen
V1 = volume of the liquid Nitrogen
V2 = volume of the gaseous Nitrogen
Calculating V2,
0.808 * 185 = 1.15 * V2
Volume of Nitrogen after expansion = 129.98 m3.
Volume = L * b * h
= 10 * 10 * 3.5
Volume of the room = 350 m3.
Fraction of air = volume of Nitrogen after expansion/volume of the room * 100
= 129.98/350 *100
= 37.14 %
Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO
Solving for X,
X = (56 g × 24.7 g) ÷ 88 g
X = 2.26 g ÷ 88 g
X = 0.0257 g of CO
Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
