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dimulka
14 days ago
10

Hydrogen has three isotopes with mass numbers of 1, 2, and 3 and has an average atomic mass of 1.00794 amu. This information ind

icates that
1. equal numbers of each isotope are present
2. more isotopes have an atomic mass of 2 or 3 than of 1
3. more isotopes have an atomic mass of 1 than of 2 or 3
4. isotopes have only an atomic mass of 1
Chemistry
2 answers:
alisha [964]14 days ago
8 0

Answer: It shows that the majority of isotopes have a mass number of 1 over 2 or 3.

Explanation:

The average atomic mass is calculated by summing the masses of each isotope weighted by their relative abundances.

The formula applied is:

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

Given three hydrogen isotopes: H-1, H-2, and H-3.

The average atomic mass is 1.00794 amu.

This value being very close to the mass of H-1 indicates its natural abundance is much greater than those of H-2 and H-3.

The abundance of H-1 is approximately 99.98%.

The combined abundances of H-2 and H-3 are about 0.02%.

Therefore, this data implies that isotopes with mass 1 are more numerous than those with masses 2 or 3.

VMariaS [1K]14 days ago
7 0

The answer is actually 3, believe me.

Explanation:

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You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation
Anarel [852]
The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.

Thus, the rotation of the racemic mixture will be equal to 0°.


- This occurs because a racemic mixture contains equal proportions of both enantiomers.
8 0
1 day ago
A chemist is studying the following reaction: NO + NO2 ⇌ N2O3. She places a mixture of NO and NO2 in a sealed container and meas
castortr0y [923]

Answer:

The forward reaction will keep occurring until all NO or all NO₂ is consumed.

Clarification:

  • According to Le Châtelier's principle, when a system at equilibrium experiences a disturbance from an outside source, the system will adjust to counteract this disturbance and restore equilibrium.

  • Thus, removing the product (N₂O₃) from the system effectively lowers the product concentration, prompting the reaction to shift forward and generate additional product in order to alleviate the strain caused by the removal of N₂O₃.

  • Consequently, the reaction will proceed forward until all of either NO or NO₂ is depleted.

5 0
2 days ago
"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92
lorasvet [956]

Answer: The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

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

From the information provided:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • For _{14}^{28}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance for _{14}^{28}\textrm{Si} isotope = 0.9222

  • For _{14}^{29}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance for _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • For _{14}^{30}\textrm{Si} isotope:

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance for _{14}^{30}\textrm{Si} isotope = x

  • The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

To convert this fractional abundance into a percentage, multiply by 100:

\Rightarrow 0.0309\times 100=3.09\%

This shows that the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

6 0
4 days ago
The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?
KiRa [971]

Answer: The mass of Si in kilograms is, 19.55kg/m^3

Explanation:

Given that the Si concentration in an Fe-Si alloy is 0.25 weight percent, this translates to:

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

Density of Fe = 7.87g/cm^3=7.87\times 10^6g/m^3

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}}}

By substituting all the provided values into this formula, we arrive at:

Si concentration in kilograms = \frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}

Si concentration in kilograms = 19.55kg/m^3

Hence, the mass of Si in kilograms is, 19.55kg/m^3

5 0
8 days ago
A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6
alisha [964]
Specific heat refers to the quantity of heat a material can absorb or release to alter its temperature by one degree Celsius. To calculate specific heat, we apply the equation for the heat absorbed by the system. The heat taken in or released by a system can be expressed by multiplying the mass of the substance by its specific heat capacity and the change in temperature. The formula is:
Heat = mC(T2-T1)
By substituting the provided values, we can find C, the specific heat of the substance.
2510 J = 0.158 kg (1000 g / 1 kg)(C)(61.0 - 32.0 °C) C = 0.5478 J/g°C
8 0
5 days ago
Read 2 more answers
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