Because its only representing the most stable isotope of that element but its considered the atomic mass and not average atomic mass
B. They are the same force. Each body acts on the other. The moon and the earth, and the apple and the earth.
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
The formula of acetic acid is CH3COOH => C2H4O2.
So, the acetic acid has the same number of atoms of carbon (C) than of oxygen (O).
Therefore, the sample that contains 96.5 moles of carbon, will contain also 96.5 moles of O.
Answer: 96.5 moles of oxygen.
The average atomic mass of tellurium, calculated from its eight isotopes (Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), Te-128 (31.79%), and Te-130 (34.48%)) is 127.723 amu.
The average atomic mass of Te can be calculated as follows:

Where:
m: is the mass
%: is the abundance percent
Knowing all the masses and abundance values, we have:

To find the <u>average atomic mass</u> we need to change all the <u>percent values</u> to <u>decimal ones</u>

Therefore, the average atomic mass of tellurium is 127.723 amu.
You can find more about average atomic mass here brainly.com/question/11096711?referrer=searchResults
I hope it helps you!