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4 years ago

How to find concentration from absorbance?

Physics

1 answer:

Alborosie4 years ago

8 0

You use Beers Law.

A = εmCl

Where A=absorbance, εm = molar extinction coefficient, C = concentration, l=path length of 1 cm)

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Is soil a product of photosynthesis?

zmey [24]

Answer:

yes

Explanation:

The plants use water (H2O) from the soil and carbon dioxide (CO2) from the air and recombine them to form carbohydrates (CH2O) and oxygen (O2).

8 0

3 years ago

Compare and contrast a series circuit with a parallel circuit. Be sure to

Flura [38]

These are my answer if you want to use them

3 0

3 years ago

If a girl running along a straight road with a uniform velocity 1.5m/s,find her acceleration

Eddi Din [679]

Her acceleration is zero, because her velocity is not changing.

6 0

3 years ago

An electronic balance shows the mass of a sample of sodium chloride to be 29.732 g. what is the uncertainty of the measurement

OLEGan [10]

The uncertainty of the measurement is 0.001 gm.

The uncertainty in the measurement of a physical quantity is given as how precisely we can measure that, in this case as we can see that the mass of the sodium chloride is precisely given as 29.732 gm, this means the electronic scale is precise to 0.001 gm and round of the values after that which means there is a uncertainty of 0.001 gm.

8 0

3 years ago

A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of th

cestrela7 [59]

Answer:

1. $a_{rad}=17545.2\frac{m}{s^{2}}$

2. $a=4429.45 \frac{m}{s^{2}}$

Explanation:

Radial acceleration is:

$a_{rad}=\frac{v^2}{r}$ (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

$v= \omega r$ (2)

By (2) on (1):

$a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)$

$a_{rad}=17545.2\frac{m}{s^{2}}$

To find the acceleration of the tube with the fall, we can use the expression:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (3)

Due impulse-momentum theorem:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (4)

with p the momentum and J the impulse. By (4) on (3):

$\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)$

And using Newton's second law (F=ma) and that (P=mv):

$mv_f-mv_i=(ma)(\varDelta t)$ (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

$\frac{mv_i}{2}=mgh$

$v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}$

So (5) is:

$-m(4.43)=(ma)(\varDelta t)$

solving for a:

$a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45$

It’s negative because is opposed to the tube movement.

5 0

3 years ago