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Hatshy [7]
3 years ago
10

What is the name for the point at which the indicator changes color in a titration? turning point end point indicator point colo

r point NextReset
Chemistry
2 answers:
Sedaia [141]3 years ago
8 0
It says on google
<span>An acid-base indicator (e.g., phenolphthalein) changes color depending on the pH. Redox indicators are also frequently used. A drop of indicator solution is added to the titration at the start; when the color changes the endpoint has been reached, this is an approximation of the equivalence point.</span>
Aloiza [94]3 years ago
8 0

Answer:

End point

Explanation:

Titration is a method of quantitative analysis which is employed to determine the concentration of unknown solutions.

Here a titrant of known concentration is added to a given volume of an analyte solution along with a suitable indicator. The titrant is added gradually until a color change in the analyte is observed. This point is known as the end point.

The end point is usually approximated as the equivalence point wherein the moles of titrant consumed equals the moles of analyte present.

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Click on the alignment to view it at a larger size. Scan along the aligned sequences, letter by letter, noting any positions whe
miss Akunina [59]

Answer:

8 amino acids

Explanation:

Amino acids are referred to as organic compounds containing carboxylic acid functional groups and amine. Amino acids contain different elements such as nitrogen, hydrogen, carbon and oxygen. Examples of amino acids are glycine,  glutamic acid and alanine. In the given alignment, there are 8 amino acids between the sequence of human and monkey.

3 0
4 years ago
Which statement is also true of the rearrangement of atoms
qwelly [4]

Answer:

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.

Explanation:

5 0
3 years ago
Read 2 more answers
What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?
Airida [17]

Answer:

V_1=0.0714L

Explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

n_1=n_2

And in terms of volume and concentration:

C_1V_1=C_2V_2

Thus, we can solve for the volume of the concentrated HCl as shown below:

V_1=\frac{C_2V_2}{C_1}

Therefore, we plug in the data to get:

V_1=\frac{0.250M*0.500L}{1.75 M}\\\\V_1=0.0714L

Best regards!

7 0
3 years ago
3. How many grams of aluminum can be heated from 90°C to 120°C if 500 J of heat energy are applied?
IgorLugansk [536]

<u>We are given:</u>

Initial Temperature = 90°c

Final Temperature = 120°c

Heat applied(ΔH) = 500 Joules

Specific heat(c) = 0.9 Joules / g°C

Mass of Aluminium(m) = ?

<u>Change in temperature:</u>

ΔT = Final temp. - Inital Temp.

ΔT = 120 - 90

ΔT = 30°c

<u>Calculating the mass:</u>

We know the formula:

ΔH = mcΔT

replacing the values:

500 = m(0.9)(30)

500 = m(27)

m = 500/27

m = 18.52 grams

5 0
3 years ago
RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
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