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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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tigry1 [53]
Density = mass/ volume
mass= 9.6 g
Volume of the object = 13.2-10.0 =3.2 ml=3.2 cm³
1ml =1 cm³

D=9.6g/3.2 cm³ = 3.0 g/cm³

3 0
3 years ago
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Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A & B B &am
Rina8888 [55]

Answer:

A) homotopic and B) enantiotopic

Explanation:

Protons chemically equivalent are those that have the same chemical shift, also if they are interchangeable by some symmetry operation or by a rapid chemical process.

The existence of symmetry axes, Cn, that relate to the protons results in the protons being homotopic, that is chemically equivalent in both chiral and aquiral environments.

The existence of a plane of symmetry, σ, makes the protons related by it, are enantiotopic and these protons will only be equivalent in an aquiral medium; if the medium is chiral both protons will be chemically NOT equivalent. The existence of a center of symmetry, i, in the molecule makes the related protons through it enantiotopic and therefore chemically only in the aquiral medium.

Diastereotopic protons cannot be interconverted by any symmetry operation and they are different, with different chemical displacement.

6 0
3 years ago
if you dilute 25.4mL of a 3.5M solution to make 166.7mL of solution, what is the molarity of the dilute solution
Anna11 [10]
Molarity=(initial molarityxinitial volume)/final volume
M=(3.5x25.4)/166.7=0.533 M
8 0
3 years ago
How many moles of Na₂CO₃ required to create 9.54 liters of a 3.4 M solution
GarryVolchara [31]

Answer:

The answer to your question is 32.44 moles

Explanation:

Data

moles of Na₂CO₃ = ?

volume = 9.54 l

concentration = 3.4 M

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

number of moles = Molarity x volume

Substitution

Number of moles = (3.4)( 9.54)

Simplification

Number of moles = 32.44

3 0
3 years ago
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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
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