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iris [78.8K]
2 years ago
12

Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount

?
Chemistry
2 answers:
shtirl [24]2 years ago
5 0
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

Murljashka [212]2 years ago
3 0
It doesn't matter what the original mass is. 1/8 is (1/2)-cubed, and 3 half-lives is 8.3 days.
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A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constantpressure calorimeter that has a heat capa
bogdanovich [222]

Explanation:

The given data is as follows.

         V_{1} = 85.0 ml,        M_{1} = 0.9 M

         V_{2} = 85.0 ml,        M_{1} = 0.9 M

Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.

So,     No. of moles = Molarity × Volume

                                = 0.9 M \times 0.085 L        (as 1 L = 1000 ml so, 85 ml = 0.085 L)

                                = 0.076 mol

As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.

              56.2 kJ/mol \times 0.076 mol

                    = 4.271 kJ

or,                 = 4271 J     (as 1 kJ = 1000 J)

Therefore,    heat released = - heat of gained by calorimeter

Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.

Hence,     mass of HCl = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Similarly,    mass of KOH = = density × Volume of HCl

                                      = 1.00 g/ml × 85.0 ml

                                       = 85 g

Hence, total mass of the solution = 85 g + 85 g

                                                        = 170 g

Also,                   q = mC \Delta T

                     4271 J = 170 g \times 325 J/^{o}C \times (T_{f} - 18.24)^{o}C    

                     0.0773 = T_{f} - 18.24

                    T_{f} = 18.317^{o}C  

Thus, we can conclude that final temperature of the mixed solution is 18.317^{o}C.

6 0
2 years ago
NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f
ELEN [110]
First, let's write down the balanced chemical reaction between the given reactants:

NO₂ + NO → N₂O + O₂

The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:

Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)

For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1

6 0
1 year ago
Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass
Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

  • 209.96 g * 22.88/100 = 48.04 g C
  • 209.96 g * 5.76/100 = 12.09 g H
  • 209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

  • 48.04 g C ÷ 12 g/mol = 4.00 mol C
  • 12.09 g H ÷ 1 g/mol = 12.09 mol H
  • 149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

3 0
1 year ago
Which object(s) formed last in our solar system?
AnnyKZ [126]

Long year ago In space, gravity attracts dust and gas together which created the young solar system. It pulled low-density cloud together to produce initially the solar nebula. These clouds are made of interstellar gas and dust.  

The sun formed first from these nebula and dust.

Planetesimals is just a process indicates the formation of Earth and the other planets from concentrations of dust and diffused matter in the solar system.

Inner planets are the planets located closure to the sun in comparison to outer planets. These inner planets are mercury, venus, earth and mars. Thus, 4.5 billion year ago Inner planets formed at last.


5 0
2 years ago
Read 2 more answers
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
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