Based on the table and the equation of reaction; the
- rate of consumption of oxygen between time 20 and 40s is 0.005 M/s
- rate of consumption of oxygen between time 40 and 60s is 0.0025 M/s
<h3>What is rate of a reaction?</h3>
The rate of a reaction is the rate at which reactants molecules are consumed or products are formed.
- Rate of reaction = change in concentration of reactants or products/ time taken
Based on the equation of reaction, the rate of consumption of oxygen is equal to the rate of consumption of methane.
Thus. rate of consumption of oxygen from time 20 seconds to time 60 seconds is calculated as follows:
Rate between 20 and 40 = 1.0 - 0.9/20
- Rate of consumption of oxygen = 0.005 M/s
Rate between 40 and 60 = 0.9 - 0.85/20
- Rate of consumption of oxygen = 0.0025 M/s
Therefore, the rate of consumption of oxygen decreases with time.
Learn more about rate of reaction at: brainly.com/question/25724512
Answer: The amino acid sequence is Leu-Ser-Val
Explanation: To produce amino acids, the DNA has to "transform" itself into a RNA by a process called Transcription. In this process, part of the DNA is transcribed into a similar RNA. The RNA produced undergoes another process, called Translation. From it, the sequence is decoded to a specific polypetide, as shown in this case. In the Translation, each 3 bases corresnpond to 1 amino acid. Because of it, the sequence above has 3 amino acids.
The N-terminal and C-terminal relates to the part where the sequence start to be read and the end of the process. Normally, the sequence starts at the 5', which correspond to the N-terminal and finishes at the 3', which matches the C-terminal. So in this example, the DNA template read 5\'GACAGACAA 3\'.
Answer:
23 moles Oxygen × 6.02 × 1023 atoms = 1.3846 × 1025 atoms Oxygen.
Explanation:
<span>As a result, the hydrogen and helium initially on these inner planets floated away into space. Only the Sun and the massive outer planets had enough gravity to keep hydrogen and helium from drifting away.</span>
Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
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