<u>Answer:</u> The average atomic mass of copper is 63.55 amu.
<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:
.....(1)
- <u>For
isotope:</u>
Mass of
isotope = 62.94 amu
Percentage abundance of
= 69.17 %
Fractional abundance of
isotope = 0.6917
- <u>For
isotope:</u>
Mass of
isotope = 64.93 amu
Percentage abundance of
= 30.83 %
Fractional abundance of
isotope = 0.3083
Putting values in equation 1, we get:
![\text{Average atomic mass of Copper}=[(62.94\times 0.6917)+(64.93\times 0.3083)]\\\\\text{Average atomic mass of copper}=63.55amu](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20Copper%7D%3D%5B%2862.94%5Ctimes%200.6917%29%2B%2864.93%5Ctimes%200.3083%29%5D%5C%5C%5C%5C%5Ctext%7BAverage%20atomic%20mass%20of%20copper%7D%3D63.55amu)
Hence, the average atomic mass of copper is 63.55 amu.
The number of sulfate ions : 2.2 x 10²³
<h3>Further explanation</h3>
Reaction
Al₂(SO₄)₃⇒2Al³⁺+3SO₄²⁻
mol Al₂(SO₄)₃(MW=342,15 g/mol) :

mol of sulfate ions (SO₄²⁻)

1 mol = 6.02 x 10²³ particles (ions, molecules, atoms), so for 0.366 mol :

Answer:
48 000 cm^2
Explanation:
8000 1cm cubes
each 1 cm cube side is 1 cm^2 and there are six sides to the cube , so each cube has 6 cm^2 surface area
and there are 8000 of them 8000 * 6 cm^2 = 48 000 cm^2
<span>The representative particle for silicon is atom of silicon.
</span>Representative particles can be atoms, molecules, formula units or ions. Representative particles depend on the nature of the substance. Silicon is chemical element made of atoms.
For sodium chloride (NaCl) for example, representative particles are ions.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)