Answer:
There are typically three ways that it is accomplished: use of erythropoietin (EPO) or synthetic oxygen carriers and blood transfusions. While transfusions of large volumes of blood or use of EPO can be detected, microdosing EPO or transfusing smaller volumes of packed red blood cells is much harder to detect.
Mass is equal to protons plus neutrons
To find the protons its the atomic number
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
Answer:
<h2>6.91 </h2>
Explanation:
To find the pH we must first find the pOH
The pOH can be found by using the formula
pOH = - log [ {OH}^{-} ]
We have
pOH = 7.09
Next we use the equation
pH = 14 - pOH
We have
pH = 14 - 7.09
We have the final answer as
<h3>6.91</h3>
Hope this helps you
Answer: The average atomic mass of the element = 88.242amu
Explanation:
The abundance of the first isotope is =35.5%
Atomic mass of first isotope = 68.9257
The average atomic mass of the first isotope =86.95amu X 35.5% =86.95amu X 0.355 =30.8725 amu
The abundance of the second isotope =64.5%
Atomic mass of the second isotope =88.95amu
The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu
Now the average atomic mass =30.8725 +57.37275 = 88.242amu
OR using the formulae
Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100
{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100
8,824/100
=88.24amu
