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forsale [732]
3 years ago
9

Match each term below with its definition or description.

Chemistry
1 answer:
zubka84 [21]3 years ago
5 0

Answer:

1. Equivalence point

2. Direct titration

3. Primary standard

4. Titrand

5. Back titration

6. Standard solution

7. Titrant

8. Indirect titration

9. End point

10. Indicator

Explanation:

1. The equivalence point is the tiration point at which the quantity or moles  of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.

2. Direct titration is a method of quantitatively determining the contents of a substance

3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance

4. A titrand is the substance of unknown concentration which is to be determined

5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration

6. A standard solution is a solution of accurately known concentration

7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution

8. Indirect titration is the process of performing a titration in athe reverse order

9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added

10 An indicator is a compound used to visually determine the pH of a solution.

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False, in all methods of charge transfer, friction, conduction, and induction, they all require at least contact between the objects.
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4 years ago
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Calculate the concentration of acetic acid and acetate ion in a 0.2 M acetate buffer at pH 5. The p K a of acetic acid is 4.76
Levart [38]

Answer:

[Acetic acid] = 0.07 M

[Acetate] = 0.13 M

Explanation:

pH of buffer = 5

pKa of acetic acid = 4.76

pH=p_{Ka} + log\frac{[Salt]}{[Acid]}

Now using Henderson-Hasselbalch equation

5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}

log\frac{[Acetate]}{[Acetic\;acid]} = 0.24

\frac{[Acetate]}{[Acetic\;acid]} = 1.74  ....... (1)

It is given that,

[Acetate] + [Acetic acid] = 0.2 M     ....... (2)

Now solving both the above equations

[Acetate] = 1.74[Acetic acid]

Substitute the concentration of acetate ion in equation (2)

1.74[Acetic acid] + [Acetic acid] = 0.2 M

[Acetic acid] = 0.2/2.74 = 0.07 M

[Acetate] = 0.2 - 0.07 = 0.13 M

5 0
3 years ago
A student uses a pH meter to measure the acidity of a water sample from a lake. For what purpose is the student MOST likely test
Mnenie [13.5K]

Answer : The correct option is, C.

Explanation :

pH meter : It is an electric device which measure the concentration of hydrogen-ion in the solution. It also measure the acidity or alkalinity of water in the solution.

Therefore, the purpose of student for testing the lake water is to understand how people use it in their daily-life. As this will help him to analyse the quality of lake water whether, it is fit for use or not.

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Answer:

Solid turns to liquid.

Explanation:

When you freeze water, it turns into ice. When melted it's still water. No chemical changes were made to the piece of ice melted. The only change was ice (solid) to water (liquid)

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5 0
3 years ago
A saturated solution of barium fluoride, BaF2BaF2, was prepared by dissolving solid BaF2BaF2 in water. The concentration of Ba2+
babunello [35]

Answer: 1.70\times 10^{-6}

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

The equation for the ionization of the BaF_2 is given as:

BaF_2\rightarrow Ba^{2+}+2F^-

When the solubility of BaF_2 is S moles/liter, then the solubility of Ba^{2+}  will be S moles/liter and solubility of F^- will be 2S moles/liter.

By stoichiometry of the reaction:

1 mole of BaF_2 gives 1 mole of Ba^{2+ and 2 moles of F^-

K_{sp}=[Ba^{2+}][F^{-}]^2

K_{sp}=s\times (2s)^2

K_{sp}=4\times (7.52\times 10^{-3})^3

K_{sp}=1.70\times 10^{-6}

Thus K_{sp} for BaF_2 is 1.70\times 10^{-6}

7 0
4 years ago
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