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rodikova [14]
3 years ago
9

The pH of pure water at 25°C is 7.0. The enthalpy change of the autoionization of water is +55.89 kJ/mol. What is the pH of pure

water at 100°C?
Physics
1 answer:
Sergeu [11.5K]3 years ago
4 0

Answer:

6.14

Explanation:

If the pH falls as temperature increases, this does not mean that water becomes more acidic at higher temperatures. A solution is acidic if there is an excess of hydrogen ions over hydroxide ions (i.e., pH < pOH). In the case of pure water, there are always the same concentration of hydrogen ions and hydroxide ions and hence, the water is still neutral (pH = pOH) - even if its pH changes.

The problem is that we are all familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that to calculate the neutral value of pH from  Kw . If that changes, then the neutral value for pH changes as well. At 100°C, the pH of pure water is 6.14, which is "neutral" on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

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A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
  • u = 48/16
  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
  • M = 0.25.

Hence, The magnification of the ornament is 0.25

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allochka39001 [22]

de mategoloalterfsqol

6 0
2 years ago
A 1kw electric heater is switched on for ten minutes. how much heat does it produce​
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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
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