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Harman [31]
3 years ago
5

Identify the titration curve for a monoprotic weak acid titrated with a strong base.

Chemistry
1 answer:
ahrayia [7]3 years ago
7 0

Answer:

Titration is a technique to determine the concentration of an unknown solution. As illustrated in the titration setup above, a solution of known concentration (titrant) is used to determine the concentration of an unknown solution (titrand or analyte).

Typically, the titrant (the solution of known concentration) is added through a burette to a known volume of the analyte (the solution of unknown concentration) until the reaction is complete. Knowing the volume of titrant added allows us to determine the concentration of the unknown analyte. Often, an indicator is used to signal the end of the reaction, the endpoint. Titrant and analyte is a pair of acid and base. Acid-base titrations are monitored by the change of pH as titration progresses.

Let us be clear about some terminologies before we get into the discussion of titration curves.

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How many energy levels does sulfur have?<br> A. 2<br> B. 3<br> C. 4<br> D. 14
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What was the initial volume of gas if a new volume of 3.00 liters is obtained when the pressure is decreased from 3.00 atm to 1.
timurjin [86]

Answer:

1 L

Explanation:

V1=P2V2/P1

Checking this, it is correct because P and V at constant T has an inverse relationship.

- Hope that helps! Please let me know if you need further explanation.

4 0
3 years ago
An unknown volume of gas has a pressure of 0.50 atm and temperature of 325 K. If the pressure is raised to 1.2 atm and the tempe
Grace [21]
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4 0
3 years ago
What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch
sammy [17]

Explanation:

It is known that charge on xenon nucleus is q_{1} equal to +54e. And, charge on the proton is q_{2} equal to +e. So, radius of the nucleus is as follows.

            r = \frac{6.0}{2}

              = 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

              d = r + 2.5

                 = (3.0 + 2.5) fm

                 = 5.5 fm

                 = 5.5 \times 10^{-15} m     (as 1 fm = 10^{-15})

Therefore, electrostatic repulsive force on proton is calculated as follows.

              F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

Putting the given values into the above formula as follows.

           F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}

              = (9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}

              = (9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}

              = 411.2 N

or,           = 4.1 \times 10^{2} N

Thus, we ca conclude that 4.1 \times 10^{2} N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0
4 years ago
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