Answer:
increase in temperature of the intrinsic semiconductor
Explanation:
- If the p-side has a higher doping concentration, it implies that number of holes (positive ion) increased which is greater than number of electron (negative ion) in the n-side
- in order to balance the intrinsic concentration, that is to balance the number of holes and electrons which depends on temperature.
- an increase in the temperature of the intrinsic semiconductor (p-side), increases the number of electron but number of holes remains constant.
A balance in the intrinsic concentration helps in tuning to the same radio channel.
The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.
<h3>What is equilibrium?</h3>
Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
A very high value of K indicates that at equilibrium most of the reactants are converted into products.
The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.
When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.
This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.
Hence, option A is correct.
Learn more about the equilibrium here:
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Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:

P₂ in terms of the ideal gas equation is:

Answer:
1) 0.18106 M is the molarity of the resulting solution.
2) 0.823 Molar is the molarity of the solution.
Explanation:
1) Volume of stock solution = 
Concentration of stock solution = 
Volume of stock solution after dilution = 
Concentration of stock solution after dilution = 
( dilution )

0.18106 M is the molarity of the resulting solution.
2)
Molarity of the solution is the moles of compound in 1 Liter solutions.

Mass of potassium permanganate = 13.0 g
Molar mass of potassium permangante = 158 g/mol
Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

0.823 Molar is the molarity of the solution.
Answer: Potential energy
Explanation: i hope i wasnt toooo late