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yan [13]
2 years ago
5

Read the passage below about acupuncture.

Chemistry
2 answers:
valkas [14]2 years ago
7 0
The answer would be "This ancient form of pain relief and stimulation is said to benefit patients by blocking channels which practitioners claim allowed pain to travel within the body" because this statement best explains how acupuncture reduces pain.

To support such answer, it must be noted that acupuncture is a traditional Chinese mode of pain-relieving through the insertion of extremely fine needles into the skin at specific points in the body known as "acupoints". This may relieve pain by releasing the hormone known as endorphins (happy hormone), which is the body's natural pain-killers and by affecting the brain part that governs serotonin- brain chemical which involved mood.
pshichka [43]2 years ago
3 0
<h3>The answer would be <u><em>"This ancient form of pain relief and stimulation is said to benefit patients by blocking channels which practitioners claim allowed pain to travel within the body"</em></u> because this statement best explains how acupuncture reduces pain.</h3><h3 />
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<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M

  • To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

pOH=14-8.08=5.92

  • To calculate pOH of the solution, we use the equation:

pOH=-\log[OH^-]

Putting values in above equation, we get:

5.92=-\log[OH^-]

[OH^-]=10^{-5.92}=1.202\times 10^{-6}M

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of X^- ions follows:

                    X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b

<u>Initial:</u>              0.1

<u>At eqllm:</u>        0.1-x                           x              x

Concentration of OH^-=x=1.202\times 10^{-6}M

The expression of K_b for above equation follows:

K_b=\frac{[OH^-][HX]}{[X^-]}

Putting values in above expression, we get:

K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:  

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.445\times 10^{-11}

Putting values in above equation, we get:

10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}

We know that:

K_a\text{ for HF}=6.8\times 10^{-6}

K_a\text{ for HCl}=1.3\times 10^{6}

K_a\text{ for HClO}=3.0\times 10^{-8}

So, the calculated K_a is approximately equal to the K_a of HF

Hence, the unknown salt is NaF

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2 years ago
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