Can you identify some techniques in separating mixtures? In what

Iridium has only two naturally occurring isotopes. Ir-191 with arelative mass of 190.96058 and Ir-193 with a relative mass of192

OLEGan [10]

Answer: The fractional abundance of Ir-191 is 0.372

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Ir-191 be x and that of Ir-193 isotope be (1-x)

For isotope 1 (Ir-191):

Mass of isotope 1 = 190.96058 amu

Fractional abundance of Ir-191 = x

For isotope 1 (Ir-193):

Mass of isotope 1 = 192.96292 amu

Fractional abundance of Ir-193 = (1 - x)

Average atomic mass of iridium = 192.217 amu

Putting values in equation 1, we get:

$192.217=[(190.96058\times x)+(192.96292\times (1-x))]\\\\x=0.372$

Hence, the fractional abundance of Ir-191 is 0.372

6 0

3 years ago

Which of the following describes what scientists expect to happen every time under certain conditions?

Ber [7]

I believe it is d. scientific law

7 0

2 years ago

Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II)

vovangra [49]

Answer:

0,051g of O₂

Explanation:

The reaction of precipitation of Fe is:

4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-

This reaction is showing you need 1 mol of O₂(g) per 4 moles of Fe(II) for a complete reaction.

85mL= 0,085L of 0,075M Fe(II) are:

0,085L*0,075M = 6,34x10⁻³ moles of Fe(II)

For a complete reaction of 6,34x10⁻³ moles of Fe(II) you need:

6,34x10⁻³ moles of Fe(II)× $\frac{1molO_2}{4moles Fe(II)}$ =

1.59x10⁻³ moles of O₂. In grams:

1.59x10⁻³ moles of O₂× $\frac{32g}{1molO_2}$ = 0,051g of O₂

I hope it helps!

7 0

2 years ago

The reaction of methane with water to form carbon dioxide and hydrogen is nonspontaneous at 298 K. At what temperature will this

KIM [24]

Answer:

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

Explanation:

CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)

CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0

ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0

S°(J/K·mol): 186.1 188.8 213.7 130.7

ΔG<0 to be spontaneous

ΔG = ΔH- TΔS <0

ΔH = ∑nΔH(products) - ∑nΔH(reactant)

ΔH = (-393.5) - (–74.87 + 2*–241.8)

ΔH = 164.97 kJ = 164970 J

ΔS = ∑nΔS(products) - ∑nΔS(reactant)

ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)

ΔS = 172.8 J

0 > 164970 J - T* 172.8 J

-164970 J > - T* 172.8 J

954.7< T

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

8 0

3 years ago

Which of an atom's electrons are involved in chemical reactions?

Nadya [2.5K]

it’s C

i hope i help you but if not DO NOT FORGET STAY SAFE AT HOME

4 0

3 years ago