Another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
<h3>What is a microspectrophotometer?</h3>
Microspectrophotometry is a biological technique used to measure the absorption or transmission spectrum of a solid or liquid material in either transmitted or reflected light.
Microspectrophotometry can also measure the emission of light by a sample, which is usually small as the micro implies.
One advantage of microspectrophotometry is that the sample does not get damaged. However,
However, another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.
Learn more about microspectrophotometry at: brainly.com/question/5832827
Answer:
3.80x1024 grams containing if not that then above 80 grams 6 grams
Explanation:
Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.
<span> 2 NaHCO</span>₂<span> </span> →<span> Na</span>₂<span>CO</span>₃<span> (s) </span>+ <span> CO</span>₂<span> (g) + H</span>₂<span>O (g)
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Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
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<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
