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zavuch27
9 days ago
12

1) Assume element X has 2 isotopes: X-125 and X-126. For every 100 atoms of X, 30 of them have a mass of 125.0 u and 70 have a m

ass of 126.0 u. What is the average atomic mass of X?
A)125.3 u B)125.7 u C)126.3 u D)126.7 u

2)What is the most reasonable inference you could make about sulfur (S), which has four stable isotopes and an average atomic mass of 32.06 u?
A). The most common isotope of sulfur has a mass of exactly 32.06 u.
B)Approximately 6% of sulfur atoms have a mass of 32 u.
C)ll of sulfur's isotopes have a mass between 32 u and 33 u.
DThe most common isotope of sulfur is S-32.
Chemistry
2 answers:
lorasvet [960]9 days ago
6 0

Answer:

Explanation:

0.3(125)+0.7(126)=37.5+88.2=125.7

2. The most prevalent isotope is 32, since the average is very close to this value.

Alekssandra [992]9 days ago
4 0

Answer:

For question 1: Option B is the correct choice.

For question 2: Option D is the right answer.

Explanation:

  • Regarding question 1:

The average atomic mass is determined by the total of the masses of each isotope, each weighted by their natural fractional abundance.

The formula used to determine average atomic mass is as follows:

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

The information provided is:

In a sample of 100 atoms of X, 30 have a mass of 125.0 u, and 70 have a mass of 126.0 u. Thus, the fractional abundance of the X-125 isotope is 0.3, while that of the X-126 isotope is 0.7.

  • For the X-125 isotope:

Mass = 125

Fractional abundance = 0.3

  • For the X-126 isotope:

Mass = 126

Fractional abundance = 0.7

Substituting the values into the equation yields:

\text{Average atomic mass }=[(125\times 0.3)+(126\times 0.7)]=125.7u

Therefore, the answer is Option B.

  • Regarding question 2:

We know:

The average atomic mass of sulfur = 32.06 u

There are 4 stable isotopes of sulfur: S-32, S-33, S-34, and S-36.

The average atomic mass of sulfur is closer to the mass of the S-32 isotope, indicating that this isotope's relative abundance is greater than that of the others.

The 'S-32' isotope is thus the most abundant isotope of sulfur.

Therefore, the correct response is Option D.

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Answer: The air in the room weighs 37.068 kg

Explanation:

Given dimensions:

Length = 10.0 ft

Width = 11.0 ft

Height = 10.0 ft

The volume of the room (rectangular prism) is calculated using:

V=lbh

where l = length, b = breadth, h = height.

Substituting the values,

V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L

Using the conversion: 1ft^3=28.3168L

Next, calculate the mass of the air based on density:

Density=\frac{Mass}{Volume}

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Identify one disadvantage to each of the following models of electron configuration:
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Answer:

Below are the downsides of each electron configuration model:

1). Dot Structures - They consume more space and fail to convey the electron distribution within orbitals.

2). Arrow and line diagrams complicate electron counting and also take up excessive space.

3). Written Configurations do not illustrate electron distribution in orbitals, leading to possible errors in counting electrons.

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Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol
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1. The luminol stock solution has a molarity of 1.431 M.

2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.

3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.

Additional Information

Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.

Key concepts in stoichiometry include:

  • 1. Relative atomic mass
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This refers to the relative atomic mass of a molecule.

  • 3. Mole

A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.

1 mole = 6.02 × 10²³ particles.

The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.

\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}

Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.

To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.

Thus, the molarity is calculated as:

  • 1. Moles of Luminol

- the relative molecular mass of Luminol:

= 8.C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams/mol.

Thus, we have:

moles = grams / relative molecular mass.

mole=\frac{19}{177}

moles = 0.1073.

2. Molarity (M)

M = moles / volume

M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}

M = 1.431.

  • b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:

moles = M × volume

moles = 6 × 10⁻² × 2

moles = 0.12.

  • c. The molarity of the stock solution (Part A) is 1.431 M.

The diluted solution (Part B) contains 0.12 moles of luminol.

To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):

volume = moles / M

volume\:=\:\frac{0.12}{1.431}

volume = 0.08386 L = 83.86 mL.

Further Learning

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Keywords: mole, volume, molarity, Luminol, relative molecular mass

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Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.

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Explanation:

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Mass of Si = 0.25 g = 0.00025 kg

Mass of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = 2.32g/cm^3=2.32\times 10^6g/m^3

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Next, we need to find the quantity of Si in kilograms per cubic meter of alloy.

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}

Si concentration in kilograms = \frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}

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