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NeTakaya
3 years ago
7

There are three naturally occurring isotopes of the hypothetical element hilarium 45Hi, 46Hi, and 48Hi. The percentages of these

isotopes are the following: Hilarium Natural Isotopes Abundance 45Hi 18.3% 46Hi 34.5% 48Hi 47.2% You can assume that the atomic masses are exactly equal to the mass number. Calculate the weight of "naturally" occurring hilarium and report it as you would for any other naturally occurring element. Answer in units of g/mol. Your answer must be within ± 0.005%
Chemistry
1 answer:
den301095 [7]3 years ago
6 0

Answer:

46.761g/mol

Explanation:

Given parameters:

Element = Hilarium , Hi

Isotopes: Hi- 45, Hi-46 and Hi- 48

Natural abundance of Hi-45 = 18.3%

                                     Hi-46 = 34.5%

                                     Hi-48 = 47.2%

Unknown:

Atomic weight of naturally occurring Hilarium = ?

Solution:

Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.

 We can use the expression below to find this value:

       Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈

    m is the atomic mass of each isotope and α is the abundance

Atomic weight = (45 x \frac{18.3}{100} ) + (46 x  \frac{34.5}{100} ) + (48 x  \frac{47.2}{100})

Atomic weight of Hi = 8.235 + 15.870 +  22.656 = 46.761g/mol

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For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH
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Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]} <em>(1)</em>

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

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Replacing in (1):

4,38x10^{-4} = \frac{x^2}{4,7-x}

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

<em>pH = 12,65</em>

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

[OH⁻] = 0,04515 - x

Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

pOH = 3,51.

<em>pH = 10,49</em>

Thus ΔpH is 12,65 - 10,49 = <em>2,16 ≈ 2,17</em>

I hope it helps!

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