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Oliga [24]
3 years ago
12

What is the major role of the Nuclear Regulatory Commission

Chemistry
1 answer:
jolli1 [7]3 years ago
5 0
The NRC<span> is headed by a five-member </span>Commission<span>. The President designates one member to serve as Chairman and official spokesperson. The </span>Commission<span> as a whole formulates policies and regulations governing </span>nuclear<span> reactor and materials safety, issues orders to licensees, and adjudicates legal matters brought before it.</span>
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Explain the general process of nuclear fission. What is created from fission?
aliya0001 [1]
In nuclear fission heavier elements are split to make lighter elements whilst releasing energy. An atom, its nucleus to be more specific, is bombarded with neutrons. The nucleus becomes unstable and it starts to split/decay. It creates the fusion products. Neutrons and lighter elements are released; the neutrons from the nuclei of the atom(s) being split.
6 0
3 years ago
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What is the mass in grams of ba(io3)2 can be dissolved in 500 ml of water at 25 degrees celcius?
lesya [120]

The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g

<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>

The Ksp of Ba(IO3)2 = 1.57 × 10^-9

Molar mass of Ba(IO3)2 = 487 g/mol?

Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

Ba(IO_{3})_{2} \leftrightharpoons Ba^{2+} + 2\:IO_{3}^{-}

Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}

[Ba(IO_{3})_{2}] =  \sqrt[3]{ksp} =\sqrt[3]{1.57 \times  {10}^{ - 9} } \\  [Ba(IO_{3})_{2}] = 1.16 \times  {10}^{-3} moldm^{-3}

moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles

mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g

Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.

Learn more about mass and moles at: brainly.com/question/15374113

#SPJ12

7 0
2 years ago
a cylinder rod formed from silicon is 46.0 cm long and has a mass of 3.00 kg. the density of silicon is 2.33 g/cm³. what is the
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d=\frac{m}{V}\\\\&#10;V=\pi r^{2}H\\d=\frac{m}{\pi r^{2}H} \ \ \ |*\pi r^{2}H\\\\&#10;d \pi r^{2}H=m \ \ \ |:d \pi H\\\\&#10;r^{2}=\frac{m}{d \pi H}\\\\&#10;r=\sqrt{\frac{m}{d \pi H}}\\\\&#10;2r=d\\\\&#10;d=2\sqrt{\frac{m}{d \pi H}}

H=43cm\\&#10;d=2.33\frac{g}{cm^{3}}\\&#10;m=3kg=3000g\\\\&#10;d=2\sqrt{\frac{m}{d \pi H}}=2*\sqrt{\frac{3000g}{2.33\frac{g}{cm^{3}} * \pi * 43cm}}=2*\sqrt{\frac{3000}{100,19\pi \frac{1}{cm^{2}}}}\approx6,17cm
7 0
3 years ago
Automobile air bags use the decomposition of sodium azide as their sources of gas for rapid inflation, represented in the reacti
Fynjy0 [20]

Answer:

The answer is 95 degree celcius

6 0
3 years ago
For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i
frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

            KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

  1. Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
  2. The charge on simple ions signifies their oxidation number.
  3. The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

                                   K + (-1) + 2(-2) = 0

                                    K-5 = 0

                                    K = +5

The oxidation number of K in KCl:

                                K + (-1) = 0

                                K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0
3 years ago
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