(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
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Answer: Heterogeneous mixture - the parts are not uniformly mixed.
A mixture contains components having distinct chemical properties. There are two types of mixtures: homogeneous and heterogeneous. In a homogeneous mixture there is uniform distribution of components. we cannot distinguish one portion of the mixture from another. for example salt mixed in water. In heterogeneous mixture, the components are not uniformly mixed. hence, we are able to distinguish different parts of a mixture, like the mixture of iron, sand and salt given in this question.
Answer:
Boyle's Law
Explanation:
Given that:
<u><em>initially:</em></u>
pressure of gas, 
volume of gas, 
<em><u>finally:</u></em>
pressure of gas, 
volume of gas, 
<u>To solve for final volume</u> 
<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>
<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>
Mathematically:
100 ml
100 ml of the stock solution is required to prepare the order.
We know that C1V1 = C2V2
where C1= 2%
V1 = 500ml
C2= 10%
V2 = ?
V2 = C1V1 / C2
= 500 * 2% / 10%
=100
V2 = 100 ml
<h3>What is meant by stock solution?</h3>
- A stock solution is a sizable amount of a typical reagent in a standardized concentration, like sodium hydroxide or hydrochloric acid.
- This phrase is frequently used in analytical chemistry while doing operations like titrations where it's crucial to employ precise solution concentrations.
<h3>What distinguishes a standard solution from a stock solution?</h3>
- The main distinction between stock solution and standard solution is that the former is a highly concentrated solution while the later is a solution whose concentration is precisely known.
- Because standard solutions frequently arrive as stock solutions, the phrases "stock solution" and "standard solution" are connected.
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