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lisov135 [29]
4 years ago
11

In an acid-base titration, a base or acid is gradually added to the other until they have completely neutralized each other. Sin

ce acids and bases are usually colorless (as are the water and salt produced in the neutralization reaction), pH is measured to monitor the reaction. Suppose that the equivalence point is reached after approximately 100 mL of a NaOH solution have been added (enough to react with all the acetic acid present) but that replicates are equally likely to indicate from 95 to 104 mL to the nearest mL. Assume that volumes are measured to the nearest mL and describe the sample space. Round your answers to one decimal place (e.g. 98.7). a) What is the probability that equivalence is indicated at 100 mL?
Chemistry
1 answer:
Xelga [282]4 years ago
5 0

Answer:

1/10 or 0.1

Explanation:

Quantities at which equivalence can be obtained are:

           

95, 96, 97, 98, 99, 100, 101, 102, 103, 104

Number of quantities = N = 10

Probability that the equivalency is obtained exactly at 100 mL can be calculated as follows:

P(X = 100) =  (Number of favourable outcomes) / (Total Number of Outcomes)

P(X = 100) = 1/10

P(X = 100) = 0.1

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ZnBr2. hopes this helps
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Read 2 more answers
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
4 years ago
How many g of MgCO3(s) are needed to make 1.2 L of 1.5 M MgCl2(aq) solution?
maw [93]
Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
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Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
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How are bones important to the lymphatic system?
Marianna [84]
Our bone marrow continuously makes new red and white blood cells. The lymphatic system consists of the bone marrow, the spleen, the thymus (in young people), and lymph nodes.
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How can you differentiate the Atomic Structure and the Bohr Model? In other words, how is the Atomic Structure different from th
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Answer:

Bohr thought that electrons orbited the nucleus in circular paths; whereas in the modern view atomic electron structure is more like 3D standing waves. Bohr built upon Rutherford's model of the atom. ... He believed that electrons moved around the nucleus in circular orbits with quantised potential and kinetic energies.

Explanation:

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