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Nesterboy [21]
3 years ago
14

Which statement best describes the difference between a substance with a pH of 3.0 and a substance with a pH of 6.0?

Physics
1 answer:
satela [25.4K]3 years ago
7 0

The statement best describes the difference between a substance with a pH of 3.0 and a substance with a pH of 6.0 is - The substance with the lower pH has 1,000 times as many hydrogen ions per volume of water.

pH is the scale or measure for the substance about its acidic or basic strength. It ranges from 0 to 6.9 which is acidic and 7.1 to 14 which is basic.

Acidic substance has high concentration of Hydrogen ions whereas, basic substance has low concentration of hydrogen ions, however for the OH⁻ ions it is reverse.

  • Concentration of hydrogen ions is inversely related to its pH
  • More hydrogen ions present, the lower the pH
  • The fewer hydrogen ions, the higher the pH We know,

pH = -log(H^{+}) then,  

=> 3 = -log (H^{+})

=> H^{+} =  (for pH = 3)  

=> pH = -log(H^{+})

=> 6 = -log (H^{+})

=> H^{+} = 10^{-6} (for pH = 6)

 Thus, The substance with the lower pH (3) has 1000 times as many hydrogen ions per volume of water.

Learn more about pH scale:

brainly.com/question/1596421

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80 x 2.5 = 200 km/hr.

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A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
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a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

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#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

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