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finlep [7]
3 years ago
5

Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.

The abundance of the middle isotope is 10.00%. Estimate the abundances of the other isotopes.
Chemistry
1 answer:
iris [78.8K]3 years ago
5 0

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

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Stolb23 [73]

Answer:

The pressure of O₂ is 0.8 atm.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case:

PT=Pnitrogen + Pcarbon dioxide + Pother gases

Being:

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  • Pcarbon dioxide: 3 mmHg
  • Pother gases: 7.1 mmHg

and replacing:

PT= 593.4 mmHg + 3 mmHg + 7.1 mmHg

you get:

PT= 603.5 mmHg

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PT= 603.5 mmHg= 0.8 atm

<u><em>The pressure of O₂ is 0.8 atm.</em></u>

6 0
3 years ago
Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0
3 years ago
A 133.8-gram sample of bronze was 10.3% tin by mass. Determine the total mass of tin in the sample.
Kay [80]
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6 0
3 years ago
The coordination compound Co3[Cr(CN)6]2 contains Co2+ cations and a complex anion. What is the likely oxidation state for Cr in
otez555 [7]

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The complex anion here is [Cr(CN)6]3-.

Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),

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7 0
3 years ago
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xenn [34]

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As 90% conjugate base is present, so propanoic acid present 10%.

8 0
3 years ago
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