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Vinil7 [7]
3 years ago
9

The amount of protein in a sample was measured by the sample's absorbance of light at a given wavelength. Using standards, a bes

t fit line (calibration curve) of absorbance (y-axes) vs. mg protein (x-axes) gave a slope of 0.0163 and an intercept of 0.104.
What is the amount of protein in an unknown sample that has an absorbance of 0.150?
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
5 0

Answer:

Protein Concentration is 2.82mg/L

Explanation:

According to Beer-Lambert's Law, Absorbance is directly proportional to the concentration.

However, the concentration of a solution can be determined from a calibration curve, in which Absorbance is plotted on the y-axis and the Concentration on the x-axis.

Plotting the best line, the equation of line is used

y = mx + c

where y is absorbance = 0.150

           m is slope = 0.0163

            x is concentration

             c is intercept = 0.104

inserting the values from the question

y = mx + c

0.150 = 0.0163x + 0.104

0.0163x = 0.150 - 0.104

0.0163x = 0.046

Divide both sides by 0.0163

0.0163x/0.0163 = 0.046/0.0163

x = 2.82

Concentration of protein = 2.82 mg/L

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The answer is 44.0095. We assume you are converting between grams CO2 and mole.
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Almost 99% of earths atmosphere is made up of two gases. What are the two gases and the percents of each?
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The Answer Is A 21% oxygen and 78%nitrogen
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Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0
3 years ago
The atomic masses of the two stable isotopes of boron; boron-10 (natural abundance:19.78%) and boron-11 (natural abundance:80.22
iogann1982 [59]
(19.78 x 10) + (80.22 x 11) all of them divided by 100= 10.81 amu
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3 years ago
How many grams of oxygen gas (02) are needed to completely react with 9.30 moles
Luba_88 [7]

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^         ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al          3 moles O₂              32.0 g
-------------------  x  ---------------------  x  --------------------  =  223 g O₂
                              4 moles Al               1 mole

6 0
2 years ago
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