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n200080 [17]
3 years ago
6

An unknown solution changed the color of a phenolphthalein indicator from colorless to pink. What would you comment about the pH

of the unknown solution? A. The given solution is acidic in nature. B. The given solution is basic in nature. C. The given solution is neutral in nature.
Chemistry
2 answers:
Brilliant_brown [7]3 years ago
7 0

Answer : The correct option is, (A) The given solution is basic in nature.

Explanation :

Indicators : It is a substance that gives the visible sign by the color change of the chemical species like an acid or a base.

Phenolphthalein, methyl orange, methyl red, etc are the type of acid-base indicators which changes the color based on pH values. They are used in the acid-base titration.

The given indicator is, Phenolphthalein.

The role of phenolphthalein indicator is that it imparts pink color to an alkaline or basic solution and it is colorless in acidic solution.

From the given problem, we conclude that the unknown solution (given solution) is, basic in nature that changes the color of a phenolphthalein indicator from colorless to pink.

Hence, the correct option is, (A) The given solution is basic in nature.

qwelly [4]3 years ago
6 0
I think the correct answer from the choices listed above is option B. <span>An unknown solution changed the color of a phenolphthalein indicator from colorless to pink. The given solution would be basic in nature. Phenolphthalein indicator is commonly used in acid-base titrations where it changes from colorless when acidic to pink when basic.</span>
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A 10 gram iron nail absorbed 125J of heat. What was the final temperature of the nail if the original temperature was 25°C?
valentina_108 [34]

Answer:

52.1 degrees C

Explanation:

We need to use the equation: q = mCΔT, where m is the mass in grams, C is the specific heat capacity, and ΔT is the change in temperature.

Here, m = 10 g and q = 125 J. The heat capacity of iron is about 0.461 J/(g * C). And, our initial temperature is 25. So:

125 J = (10 g) * (0.461 J/(g * C)) * (T_f - 25)

Solving for T_f (final temp), we get: 52.1 degrees C

Hope this helps!

7 0
4 years ago
Where do elements in the human body tend to be located on the PERIODIC TABLE? (NOT PERCENTAGES PLEASE)​
Ghella [55]

Answer:

Oxygen, Carbon, Hydrogern, Nitrogen, Sulfur, and Chlorine are all non-metals

Calcium, Potassium, Sodium, and Magnesium are all metals

Explanation:

8 0
3 years ago
Which of the following is a statement of Hess's law?
amid [387]

Answer:

Letter C...........................

Explanation:

Letter C

5 0
3 years ago
Read 2 more answers
300 grams of a compound which contains only carbon, hydrogen and oxygen is analyzed and found to contain the exact same percenta
ser-zykov [4K]

The empirical formula of the compound is C₄H₅O₃.

<u>Explanation</u>:

If hydrogen % = 5.98823, the Sum of C+O = 94.01177% or 47.005885% each

The original mass of 300 g is not important

C = 47.005885 / 12.011 = 3.913569

H = 5.98823 / 1.008 = 5.94070  

O = 47.005885 / 15.999 = 2.93805  

Divide by the smallest number:      

C =1.33  

H = 2.022

O = 1

Multiply through by 3

C = 4

H =6

O = 3

The empirical formula of a compound = C₄H₆O₃

Do not be confused by the 300 g. It is totally irrelevant to the question because you are dealing with % amounts. It would only have been of importance if you were given some mass values.

4 0
3 years ago
238 U 92 write the following decay sequence.​
Vladimir [108]

Answer: The given decay sequence is ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta.

Explanation:

An alpha-particle is a helium atom. Hence, when an alpha decay occurs in ^{238}_{92}U then the reaction equation is as follows.

^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He

Now, in sequence the equation for beta decay is as follows.

^{234}_{90}Th \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta

Hence, the sequence will be as follows.

^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta

Thus, we can conclude that the given decay sequence is ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta.

4 0
3 years ago
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