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LekaFEV [45]
4 years ago
8

In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed

even though it was shaded from the sun by a dark cloth. At first, how did Becquerel explain this result?
Chemistry
1 answer:
marshall27 [118]4 years ago
3 0
<span>In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed even though it was shaded from the sun by a dark cloth. The explanation that he gave was that the crystals were emitting light without any external light source to stimulate or excite them. He called this phenomenon as "radioactivity'". Further, more and more experiments were done to analyse actually what happened between the crystals and the light</span>
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A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
A sample of a gas in a rigid container has an initial pressure of 1.049 kPa and an initial temperature of 7.39 K. The temperatur
Doss [256]

Answer:- 4.36 kPa

Solution:- At constant volume, the pressure of the gas is directly proportional to the kelvin temperature.

\frac{P_1}{T_1}=\frac{P_2}{T_2}

Where the subscripts 1 and 2 are representing initial and final quantities.

From given data:

P_1 = 1.049 kPa

P_2 = ?

T_1 = 7.39 K

T_2 = 30.70 K

For final pressure, the equation could also be rearranged as:

P_2=\frac{P_1T_2}{T_1}

Let's plug in the values in it:

P_2=\frac{1.049kPa(30.70K)}{7.39K}

P_2 = 4.36 kPa

So, the new pressure of the gas is 4.36 kPa.

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How is heat transferred when a person holds an ice pack?
Darina [25.2K]
I’m pretty sure it’s b.
8 0
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¿Cuál es el %m/v de azúcar, en una disolución que se prepara disolviendo 14 g de azúcar en agua, para hacer 250 ml de disolución
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Answer:

The answer is  354 g.Explanation:

6 0
3 years ago
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That would be c, the alveoli.
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