What is the molarity of potassium ions in a 0.122 M K2CrO4 solution
2 answers:
Answer : The molarity of potassium ions in a solution is 0.244 M
Explanation :
The dissociation reaction of in solution will be:
From the reaction we conclude that, 1 mole of dissociate in solution to give 2 moles of
ion (potassium ion) and 1 mole of
ion (chromate ion).
As per question,
0.122 M of dissociate in solution to give
of
ion (potassium ion) and 0.122 M of
ion (chromate ion).
Hence, the molarity of potassium ions in a solution is 0.244 M
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The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
</span>
To calculate this, we will use two equations:
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>
</span>
It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>
</span><span>
</span>
Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
Answer:
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is
Explanation:
Given :
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:
= Rydberg energy
n = principal quantum number of the orbital
Energy of 11th orbit =
Energy of 10th orbit =
Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.
(Planck's' equation)
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is
Hence the resulting concentration of Li+ is 1M. Hope it helps.
