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Paraphin [41]
3 years ago
14

How are we supposed to answer this question?​

Chemistry
1 answer:
marysya [2.9K]3 years ago
6 0

Answer:

Stoichiometry

Explanation:

You have to use the stoichiometry rules. Remember that concentracion is the same as moles divided by the volume in the system

C = n/V

Also, that amount of moles is equal to mass divided by the molarmass

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Thulium-167 has a half life of 9.0 days. If the initial amount of thulium is 50.0 grams. How many grams are left after 36.0 days
BARSIC [14]

Answer:

3.125 grams.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

∵ Thulium-167 has a half life of 9.0 days.

∴ The time is needed to calculate the grams are left after (36.0 days) represents (36.0 days / 9.0 days) = 4.0 half-lives.

50.0 grams → (first half life) 25.0 grams → (second half life) 12.5 grams → (third half life) 6.25 grams → (fourth half life) 3.125 grams.

So, the grams are left after 36.0 days = 3.125 grams.

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3 years ago
Which of the following would a chemist be most likely to study?
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B. because the other options are different sciences (biology and entomology)
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3 years ago
A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib
rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of Ni^{2+} ions at equilibrium is, 4.31\times 10^{-8}

Explanation :  Given,

Moles of NiCl_2 = 0.130 mol

Volume of solution = 1 L

Concentration=\frac{Moles }{Volume}

Concentration of NiCl_2 = Concentration of Ni^{2+} = 0.130 M

Concentration of NH_3 = 1.20 M

K_f=5.5\times 10^8

The equilibrium reaction will be:

                      Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}

Initial conc.    0.130       1.20                     0

At eqm.             x       [1.20-6(0.130)]      0.130

                                   = 0.42

The expression for equilibrium constant is:

K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}

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

5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}

x=4.31\times 10^{-8}

Thus, the concentration of Ni^{2+} ions at equilibrium is, 4.31\times 10^{-8}

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Nuclear reactions involves​
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