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Karolina [17]
3 years ago
8

Iodine-131 is a radioactive isotope. after 9.00 days, 46.0% of a sample of 131i remains. what is the half-life of 131i?

Chemistry
1 answer:
4vir4ik [10]3 years ago
6 0

For this problem, we use the integrated rate law for first order radioactive decay which is expressed as follows:<span>

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant. 

We need to calculate first the value of k from the given ration An/Ao and the time it reached that ration. We do as follows:

An = Aoe^-kt
0.46= e^-k(9)
k = 0.0863 / day

At half life, the remaining substance would be equal to one-half of the original so that An/Ao would be equal to 1/2 or 0.50. We calculate the half-life as follows:</span>

<span>
An = Aoe^-kt
0.50 = e^-0.0863(t)
<span>t = 8.03 days</span></span>

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A decomposition reaction has a half-life that does not depend on the initial concentration of the reactant.what is the order of
kupik [55]
The reaction is first order. 
7 0
3 years ago
Question 3 (Molarity)
antiseptic1488 [7]

Answer:

The molarity of the solution is 1,03 M.

Explanation:

Molarity is a concentration measure that expresses the moles of solute (in this case HBR) in 1 liter of solution (1000ml). First we calculate the mass of 1 mol of HBr, to calculate the moles that are in 50 g of said compound:

Weight 1 mol HBr= Weight H + Weight Br= 1,01g + 79,90g= 80, 91 g/mol

80,91 g ----1 mol HBr

50,0 g------x= (50,0 g x1 mol HBr)/80,91 g= 0,62 mol HBr

600 ml solution-----0,62 mol HBr

1000ml solution------x= (1000ml solution x 0,62 mol HBr)/600 ml solution

<em>x=1,03 moles HBr ---> The solution is 1,03M</em>

8 0
3 years ago
What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
What are the four symbols for physical states of reactants and products?
Studentka2010 [4]
Solid -(s)
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gas - (g)
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7 0
3 years ago
An element has an atomic number of 18 and an atomic mass of 40. The number of neutrons in the nucleus of an atom of this element
Vikentia [17]

Answer: 22 neutrons

Explanation: 40 is the mass number = atomic mass = total number of protons and neutrons in atomic nucleus

18 is the number of protons in the nucleus of this atom

Then 40 - 18 = 22 neutrons

and this is Argon

7 0
3 years ago
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