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stellarik [79]
3 years ago
6

Uranium (u235), naturally occurring at 0.7%, must be ____________ before it can be put into a reactor. enriched to 3% extensivel

y processed in order to change it to a nonreactive form produced by bombarding it with electrons from radioactive plutonium washed to lower its concentration
Chemistry
1 answer:
Ierofanga [76]3 years ago
4 0

Answer: -

Extensively processed to clean, purify, and concentrate it.

Uranium (U235), naturally occurring at 0.7%, must undergo a series of processes to become a useable fuel before it can be put into a reactor.

The process include converting uranium oxide to uranium hexafluoride allowing it to be enriched to 3%.

After enrichment, the gas is converted to uranium dioxide fuel pellets. These are placed inside thin metal tubes which are assembled in bundles to become the fuel elements for the core of the reactor.

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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.

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Considering the following precipitation reaction: Pb(NO3)2(aq) + 2Kl(aq) – Pbl (s) + 2KNO3(aq) IN What is the correct complete i
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Answer:

Pb²++2I−→PbI2(s)

yellow precipitate

Explanation:

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What class of elements includes all of the elements that are gases at room temperature? nonmetals metals metalloids?
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Metals have high density. They are always solids at room temperature.
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Non-metals have low density. They can be solids, liquids or gases at room temperature.

Based on this, the <span>class of elements that includes all of the elements that are gases at room temperature is the non-metals</span>
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solution:

You need to find the frequency, and they have already given you the wavelength. And since you already know the speed of light, you can use formula (2) to answer this problem. Remember to convert the nano meters to meters because the speed of light is in meters. (1 nm = 1.0 x 10^{-9} m)

v=c\lambda \\v= (3.00 \times 10^8 m/s)/(6.9\times 10^-7 m) \\v = 4.35 \times 10^{14} s^{-1}

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