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zalisa [80]
3 years ago
6

Write a nuclear equation to describe the spontaneous fission of 95 244 Am to form I-134 and Mo-107. Determine how many neutrons

are produced in the reaction
Chemistry
1 answer:
MissTica3 years ago
4 0

<u>Answer:</u> 3 neutrons are produced in the above nuclear reaction.

<u>Explanation:</u>

In a nuclear reaction, the total mass and total atomic number remains the same.

The nuclear reaction for the fission of Americium-244 isotope follows:

^{244}_{95}\textrm{Am}\rightarrow ^{134}_{53}\textrm{I}+^{107}_{42}\textrm{Mo}+Y^1_0\textrm{n}

<u>To calculate Y:</u>

Total mass on reactant side = total mass on product side

\Rightarrow 244=134+107+(Y\times 1)\\\\\Rightarrow Y=244-241=3

Hence, 3 neutrons are produced in the above nuclear reaction.

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A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. T
AveGali [126]

Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

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Answer:

\rm 2\; I^{-} + Cl_2 \to I_2 + 2 \; Cl^{-}.

Start color: yellowish-green.

End color: dark purple.

Assumption: no other ion in the solution is colored.

Explanation:

In this reaction, chlorine gas \rm Cl_2 oxidizes iodine ions \rm I^{-} to elemental iodide \rm I_2. At the same time, the chlorine atoms are converted to chloride ions \rm Cl^{-}.

Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.

Among the reactants of this reaction, \rm I^{-} is colorless. If there's no other colored ion, only the yellowish-green hue of \rm Cl_2 would be visible. Hence the initial color of the reaction would be the yellowish-green color of \rm Cl_2.

Similarly, among the products of this reaction, \rm Cl^{-} is colorless. If there's no other colored ion, only the dark purple hue of \rm I_2 would be visible. Hence the initial color of the reaction would be the dark purple color of \rm I_2.

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