What should the absorbance be if light is passed through the same solution at the same wavelength and the light path is reduced to 1 cm?

.20 is the correct answer to this question. The type of question that is being asked is a math word problem. With math word problems, a person reading it will need to use the words in the problem to then make a formula from that.

The two key words in this math word problem are wave length and absorbance. Using those words, we would know that the formula would include the absorbance amount and the wavelength.

So to properly answer this question, we would divide .40, by the 2 centimetres light path. Doing this division will give the correct answer of .20.

The correct answer to this question is 0.20. This is a math word problem where a formula should be used to solve the problem. In this math word problem, the formula should include the fact that the absorbance amount is the same as the molar absorptivity amount. This is then multiplied by a pathlength.

That total should then be multiplied by the amount of concentration. When this formula is created, on the left side of the formula, the answer will be .40. Then, when plugging the rest of the formula in, the answer will be .20 as .40 is divided by 2

You would be using the Beer−Lambert law for this. Each wavelength of light that is going to pass through the spectrometer will have different intensities. There is a chart available that will allow you to know different transmittance and absorbance of the different wavelengths of life.

For example, if the transmittance is 15%, expect that the absorbance is going to be 0.824 AU. If the transmittance is going to be 10%, that is the time when the absorbance is going to be 1. The transmittance rate is going to make a lot of difference with the absorbance. The lower the transmittance rate, the higher the absorbance rate is going to be.

The best way to answer this question is by using a formula. The formula to use here is Absorbance= eLc (where e is the molar extinction coefficient, L is the path length, and c is the concentration of the solution). However, to answer this question, you need to understand Beer-Lambert's law.

According to Beer-Lambert's law, Absorbance is always proportional to the path length, provided the concentration of the solution and the molar extinction coefficient is kept constant. This is quite similar to this question, as both the concentration of the solution and the molar extinction coefficient are kept constant.

Since the value of the path length is reduced by half (from 2 to 1), the value of the absorbance will be reduced by half. This means if the value of the absorbance was recorded as 2.0, then it will be reduced by half to give 1.0.

In a math word problem, it is common to use a formula. In this case when the math word problem includes absorbance and wave length, then the formula should be used where the absorbance amount is equal to the amount of the molar absorptivity multiplied by the pathlength and multiplied that by the concentration amount. With that said, the problem gives the amount being absorbed, so the answer on the left of the equal side is given at 0.40. The light path is 2 centimeters and it is reduced to 1 cm.

When plugging in the numbers, the answer will come out to be 0.20 based on having to divide 0.40 by the light path amount of two after moving some of the numbers around to get the variable on one side.