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balu736 [363]
3 years ago
11

A pH change can be evidence that

Chemistry
1 answer:
olganol [36]3 years ago
7 0

Answer:

Explanation:The pH scale ranges from 0 to 14. The pH of a solution is a measure of its acidity or alkalinity (base). ... A change of one unit on the pH scale represents a change in the concentration of hydrogen ions by a factor of 10, a change in two units represents a change in the concentration of hydrogen ions by a factor of 100.

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draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum
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Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

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Someone please help will mark as brainliest
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Which of the following describes the physical state of the layers of the Earth from the outside to the center?
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The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho
sp2606 [1]

Explanation:

(a)   It is known that relation between energy and mass is as follows.

            E = 2 \times mc^{2}

where,    E = energy

              m = mass

              c = speed of light = 3 \times 10^{8} m/s

As it is given that mass is 9.109 \times 10^{-31} kg. So, putting the given values into the above formula as follows.

             E = 2 \times mc^{2}

                       = 2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s

                       = 1.638 \times 10^{-13} J

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is 1.638 \times 10^{-13} J.

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

                    E = h \times \nu   ..... (1)

where,     h = plank's constant = 6.626 \times 10^{-34} J.s

              \nu = frequency

Also,     E = 2 \times mc^{2} ........ (2)

Hence, equating equations (1) and (2) as follows.

                    h \times \nu = 2 \times mc^{2}        

So,    

6.626 \times 10^{-34} Js \times \nu = 1.638 \times 10^{-13} J

                           \nu = 1.236 \times 10^{20} Hz

Thus, we can conclude that the frequency is 1.236 \times 10^{20} Hz.

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