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2 years ago

When a hydrocarbon burns in air what component of air reacts?

Chemistry

1 answer:

andre [41]2 years ago

5 0

Answer:

oxygen level reduces

Explanation:

because oxygen supports burning

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Normal red blood cells slide easily through narrow blood vessels. In sickle cell disease, many red blood cells change to a cresc

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Answer:

C. The child must have inherited sickle cell genes from both parents.

Explanation:

Because of the fact that sickle cell disease is recessive, the child would have to get the gene from both parents.

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2 years ago

Which energy transformation takes place in a gasoline lawn mower?

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A lawn mower works by converting chemical energy (energy stored in the gas) to mechanical energy.

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3 years ago

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Which two elements have the most similar chemical properties? 1. beryllium and magnesium 2. hydrogen and helium 3. phosphorus an

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3 years ago

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Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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 $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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2 years ago