<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 correct option is:
The latent heat of vaporization of water is given by option D. 22,590J
Latent heat of vaporization:
Latent heat of vaporization is energy required to change 1 kg of liquid to gas at atmospheric pressure at its boiling point.
Also, Latent heat is the energy released or absorbed during a phase change.
Energy released during phase change is Exothermic process.
Whereas energy absorbed during phase change is Endothermic process.
Q=mL
Q= Latent heat
m= mass= 10 g
L= Latent heat of vaporization= 2,259 J/g
Putting the values in the above equation:
Q=10*2259
Q=22590 L
Learn more about 1st law of thermodynamics here,
brainly.com/question/20114432
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Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>