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sdas [7]
3 years ago
5

Determine the percent composition of Ca3(PO4)2

Chemistry
1 answer:
Tamiku [17]3 years ago
8 0

determine the percent of composition

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5 0
2 years ago
In a reaction, chlorine accepts an electron to form Cl-. What is true about the atomic size of the chlorine ion?
Lynna [10]

Answer is: The atomic size of the chlorine ion is larger than the size of the chlorine atom.

Covalent radii of chlorine atom (Cl) is 0.099 nm and ionic radii of chlorine anion (Cl⁻) is 0.181 nm.

Difference between an chlorine atom and chlorine anion is the number of electrons that surround the nucleus.

Chlorine atom has 17 electrons and chlorine anion has 18 electrons.

6 0
3 years ago
Read 2 more answers
Iron (III) oxide pentahydrate formula
Nina [5.8K]

Answer:

Molecular Formula Fe2O12S3·5H2O

IUPAC Name iron(3+);tri sulfate;pentahydrate

Explanation:

4 0
3 years ago
How many atoms are there in one molecule of C22H44O?
Alik [6]

Answer:

64

Explanation:

8 0
3 years ago
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The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C
Kay [80]

<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

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'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

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

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

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

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

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

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.775\times 100=77.5\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.775)=0.225\times 100=22.5\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

6 0
3 years ago
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