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bogdanovich [222]
3 years ago
6

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

are the average percent abundances of each isotope?
Chemistry
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

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0.32 M

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We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Keq = [Ag⁺]² × [S²⁻]

[Ag⁺] = √{Keq / [S²⁻]}

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8 0
3 years ago
Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with
Sphinxa [80]

Answer : The work done is, 1.98\times 10^4J

Explanation :

The given balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

When 4 moles of N_2 react with 12 moles of H_2 then it gives 8 moles of NH_3

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of N_2 + Moles of H_2

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of NH_3

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 25^oC=273+25=298K

Putting values in above equation, we get:

1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K

V=195.7L

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V

where,

w = work done

p = pressure of the gas = 1.0 atm

\Delta V = change in volume = -195.7 L

Now put all the given values in the above formula, we get:

w=-p\Delta V

w=-(1.0atm)\times (-195.7L)

w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, 1.98\times 10^4J

6 0
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