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babymother [125]
3 years ago
12

It is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The n

uclide captures a neutron to form , which in turn undergoes β decay. The daughter nuclide produces the characteristic γ rays used for the analysis. What is the daughter nuclide?

Chemistry
1 answer:
Artyom0805 [142]3 years ago
5 0

Answer:

The daughter nuclide is selenium; 76:34 Se

Explanation:

The complete question is as follows;

is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide 75:33 As captures a neutron to form 76:33 As, which in turn undergoes beta decay. The daughter nuclide produces the characteristic gamma rays used for the analysis. What is the daughter nuclide?

solution:

Please check attachment for decay equations and explanations

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Which type of molecule forms structures in cells and stores energy? (5 points)
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A square painting has a length of 62 cm.<br><br> What is the area of the painting?<br><br> cm²
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Explanation:

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 A chemist describes a particular experiment in this way: "0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol o
AleksandrR [38]
Description:

<span>"0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol of O2." 

This means that a certain amount of H2O2 (0.0400 mol) decomposed or was broken down into two components, 0.04 mol of H2O and 0.02 mol of O2. To examine the system, we need a balanced equation:

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4 0
3 years ago
Given the unbalanced equation below, answer the following: Calculate the number of liters of 3.00 M lead (II) iodide solution pr
mr_godi [17]

The number of liters of 3.00 M lead (II) iodide : 0.277 L

<h3>Further explanation</h3>

Reaction(balanced)

Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)

moles of KI = 1.66

From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

\tt \dfrac{1}{2}\times 1.66=0.83

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.83}{3}\\\\V=0.277~L

5 0
3 years ago
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