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3 years ago

What nuclide would be produced from the beta particle emission of Np-236?

1 answer:

4 0

Neptium-236 has 143 neutrons and a half-life of 154,000 years. It can decay ny electron capture, beta emission and alpha emission. For beta emission, the decay product is Plotunium-236. Hope this answers the question. Have a nice day.

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1 A pump with an 80% efficiency drives water up between two reservoirs through a piping system of total length L = 15 and circul

LenKa [72]

The correct answer to this question is that the length of 14 is it’s half Which would be 7

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2 years ago

Describe the three subatomic particles (protons, neutrons, and electrons) in terms of their mass, charge, and location within th

atroni [7]

Particle mass charge      location

protons    1.673*10 ^ - 27 kg 1.6*10 ^ -19 C in the nucleus
neutrons   1.675*10 ^ - 27 kg 0    in the nucleus
electron    9.11 * 10 ^ - 31 kg    -1.6 * 10 ^ - 19 C    around the nucleus (orbitals)

Protons and neutrons have almost same masses. Mass of electrons is 1/1840 the mass of the protons.

Protons and electrons have the same magnitud of charge with different sign. Protons are positive and electrons are negative. Neutrons do not have charge.

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3 years ago

Which element has the smallest atomic radius? Fluorine, Iron, Phosphorous, Potassium

Ivahew [28]

Answer:

actually answer should be helium

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3 years ago

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vova2212 [387]

D. The African Plate

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3 years ago

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

3 years ago