Answer:
d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.
Explanation:
This is an example of a n-type semiconductor. The additional electron introduced to the 'grid' of silicon atoms causes an increase in the conductivity of the silicon. This additional electron is introduced as arsenic loses its extra electron.
Answer : Option A) Translation
Explanation : A composition of reflections over parallel lines is the same as a <u>Translation.</u>
To identify if the composition of reflections over parallel lines are same as translation or not?
We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.
Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.
PV = nRT
V, T and R are constants so they can be lumped together to a constant k.
P/n = k
P1/n1 = P2/n2
since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
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
