Does granite have cleavage or fracture?

What is the kernel form of a ferric ion? (Electron Configuration)

AnnZ [28]

1s2, 2s2, 2p6, 3s2, 3p6, 3d5

3 0

3 years ago

What is this element? 1s22s22p63s2 Magnesium Aluminum Potassium Neon

valkas [14]

Answer:

It should be silicon... I even googled it to see if I was correct and that’s what it says so can someone figure out why silicon is not an answer

Explanation:

6 0

2 years ago

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what is the total amount of energy required to remove the most loosely bound electron from each atom in a mole of gaseous Ca

Temka [501]

Answer:

589.8 kJ/mol

Explanation:

Let's consider the process in which the most loosely bound electron from each atom in a mole atom of gaseous calcium is removed.

Ca(g) + Energy ⇒ Ca¹⁺(g) + 1 e⁻

The energy involved in this process is known as First Ionization Energy. The First Ionization Energy for calcium is 589.8 kJ/mol.

5 0

3 years ago

What type of reaction is

Alex17521 [72]

Answer:

You have a chemical reaction that is occurring

8 0

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.

6 0

3 years ago