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oksian1 [2.3K]
2 years ago
11

Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

an isotopic mass of 87.0 atomic mass units, the average atomic mass of element X is numerically equal to
Chemistry
1 answer:
bija089 [108]2 years ago
8 0
<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic  \: mass }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 72 \times84.9 + 28 \times 87  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 6112.8 + 2436  }{100} \\

\footnotesize \longrightarrow \:  \rm Average \:  atomic  \: mass =  \dfrac{ 8548.8  }{100} \\

\footnotesize \longrightarrow \:  \bf Average \:  atomic  \: mass =  85.488 \: amu  \\

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Answer:

The ΔG° is 29 kJ and the reaction is favored towards reactant.

Explanation:

Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,  

ΔG° = ΔH°rxn - TΔS°rxn

= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K

= 41.2 kJ - 12.2 kJ

= 29 kJ

As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.  

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3 years ago
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The answer is: It's solubility
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For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac
barxatty [35]
First, we need to get the molar mass of:

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KCl =39.1 + 35.5 = 74.6 g/mol

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From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3: 

when 245.2 KClO3 gives → 96 g  O2

   2.72 g KClO3 gives →  X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

     0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

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C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

       83.6 Kg KClO3 gives  →  X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

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D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

        22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

      = 8.8 mg of O2

     

 


7 0
3 years ago
Can someone help me with this molar mass problem?[It’s the last one]
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Answer:

54.18 \times 10^{23} \ moles in 3 mole of  Al_2(SO_4)_3

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Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Since 1 ion of anything is equivalent to 6.02\times10^{23} \ moles

Therefore 3 moles of Al_2(SO_4)_3 will have 3\times3=9 \ ions of   SO_4^2^-

Which is equivalent to 9 \times6.02\times10^{23}=54.18\times10^{23} \ moles

Thus 3 moles of  Al_2(SO_4)_3 gives 54.18\times10^{23} \ moles of  SO_4^2^-.

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