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Arte-miy333 [17]
3 years ago
13

E=hv h= 6.62 x 10^-34 m^2 kg/ s

Chemistry
1 answer:
Drupady [299]3 years ago
7 0

v=5.0×10^12Hz

\\ \sf\longmapsto E=hv

\\ \sf\longmapsto E=6.626\times 10^{-34}Js\times 5\times 10^12s^{-1}

\\ \sf\longmapsto E=33.13\times 10^{-22}J

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Select all steps below that you followed to practice laboratory safety during the experiment. used tongs or a test tube holder t
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Explanation:

Steps followed to practice laboratory safety during the experiment are as follows.

  • Used tongs or a test tube holder to hold materials over the Bunsen burner flame.
  • Wore gloves and goggles.
  • Made careful observations of the products and reactants.
  • Did not smell the gases produced.

When we heat a test tube over bunsen flame then the tube gets hot and when we hold it with bare hands then out hands will burn. Therefore, it is advised to hold test tube with the help of tongs or a holder so that our hands did not burn.

We should also wear gloves and goggles so that any acid would not spill directly on our hands, skin and eyes as it can affect or damage the skin severely.

Careful observations were made so that correct calculations about the experiment can be carried out.

It is also advised that we should not smell the gases produced but gases move freely from one place to another in a laboratory or any where else.

So, we can try to avoid it by covering our mouth with a cloth but we cannot stop it. If we keep on inhaling the gases produced in a laboratory then it can also lead to severe disease or defect in the human body.

8 0
3 years ago
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All matter has <br><br>A. has mass<br>B. is atteacted to all other matter​
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Answer:

a

Explanation:

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3 years ago
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What is the concentration (M) of sodium ions in 4.57 L of a 0.398 M Na3PO4 solution?
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For the titration of a weak acid with a strong base what is the pKa of the weak acid if the pH is 6.72 at the equilvalence point
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Explanation:

We have to calculate pK_{a} value.

It is known that at the equivalence point concentration of acid is equal to the concentration of anion formed.

Hence,         [HA] = [A^{-}]

Now, relation between pK_{a} and pH is as follows.

                   pH = pK_{a} + log \frac{[A^{-}]}{[HA]}

Putting the values into the above formula as follows.

                   pH = pK_{a} + log \frac{[A^{-}]}{[HA]}

                  4.23 = pK_{a} + log (1)            (as [HA] = [A^{-}])

                   pK_{a} = 4.23                (as log (1) = 0)

or,                   pK_{a} = 4

Thus, we can conclude that pK_{a} of given weak acid is 4.  

                   

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Answer:

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Explanation:

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