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aleksandr82 [10.1K]
3 years ago
11

liam uses a machine that counts waves on a metal cable. The machine counts 25 waves in 2 seconds. what is the frequency of the w

ave?
Chemistry
1 answer:
Softa [21]3 years ago
7 0
Frequency is defined as the number of waves per second. In this machine 25 waves pass in one second.
We need to calculate the number of waves that pass a particular point during one second.
During 2 seconds -25 waves
Therefore in one second - 25/2 = 12.5 waves/s.
1 wave per second has the unit Hertz (Hz)
Therefore answer is 12.5 Hz
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Four or Five

Explanation:

The five phases of matter. There are four natural states of matter: Solids, liquids, gases and plasma. The fifth state is the man-made Bose-Einstein condensates

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Que son los metaloides y que caracteristicas poseen
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crash course video about periodic table - The Periodic Table: Crash Course Chemistry #4 - write a reflection about this video
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If objects with a greater mass have greater gravitational pull, why is our solar
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It's too far away

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The neutralization of H 3 PO 4 with KOH is exothermic. H 3 PO 4 ( aq ) + 3 KOH ( aq ) ⟶ 3 H 2 O ( l ) + K 3 PO 4 ( aq ) + 173.2
Ratling [72]

Explanation:

It is known that,

      No. of moles = Molarity × Volume

So, we will calculate the moles of H_{3}PO_{4} as follows.

         No. of moles = 0.227 \times 0.055 L

                                = 0.0125 mol

Now, the moles of KOH are as follows.

       No. of moles = 0.680 \times 0.055 L

                                = 0.0374 mol

And, 3 \times \text{moles of} H_{3}PO_{4} = 3 \times 0.0125

                             = 0.0375 mol

Now, the balanced reaction equation is as follows.

     H_{3}PO_{4}(aq) + 3KOH(aq) \rightarrow 3H_{2}O(l) + K_{3}PO_{4}(aq) + 173.2 kJ

This means 1 mole of H_{3}PO_{4} produces 173.2 kJ of heat. And, the amount of heat produced by 0.0125 moles of H_{3}PO_{4} is as follows.

         M = \frac{0.0125 mol \times 173.2 kJ}{1}

             = 2.165 kJ

Total volume of the solution = (55.0 + 55.0) ml

                                               = 110 ml

Density of the solution = 1.13 g/ml

Mass of the solution = Volume × Density

                                  = 110 ml \times 1.13 g/ml

                                  = 124.3 g

Specific heat = 3.78 J/g^{o}C

Now, we will calculate the final temperature as follows.

              q = mC \times \Delta T

          2165 J = 124.3 \times 3.78 \times (T - 22.62)^{o}C

        2165 - 469.854 = T - 22.62^{o}C

           17.417 = T - 22.62^{o}C

               T = 40.04^{o}C

Thus, we can conclude that final temperature of the solution is 40.04^{o}C.

3 0
3 years ago
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