Hi there, your question was incomplete, so I will be solving a similar question so it would help you to understand how to answer these types of questions
Q) A chemist must dilute 22.5mL of 119μ M aqueous silver(II) oxide AgO solution until the concentration falls to 65.0μ M . He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to significant digits.
Answer:
<u>Final Volume V2 = 41.2 mL</u>
Explanation:
We will start by first recalling the formula of dilution which goes by,
M1 * V1 = M2 * V2 ------------------------------(1)
Now let us consider V1 = 22.5 mL
M1 = 119μ M
M2 = 65.0μ M
lets substitute the values above in equation 1,
M1 * V1 = M2 * V2
V2 = (M1 * V1)/M2 ( I have created V2 as the subject here)
V2 = (119*22.5)/65
<u>V2 = 41.2 mL</u>
Answer:
10seconds
Explanation:
use the formula a=v final_v inital/time
Answer:
When light enters from air to water i.e. it is moving from rarer to denser medium, it changes its original path as there is a change of speed of light and deflects itself towards the normal. This is known as the refraction of light and this is why a pencil in a cup of water looks as if it is broken and larger.
Explanation:
Answer:
Part A: D
Part B: W = Qh - Qc
Part C: e = 1 - Qc/Qh
Explanation:
The heat engine is the engine that transforms heat (Q) in work (W), and by the second law of the thermodynamics, its efficiency can not be 100%, it means that some heat must be dissipated.
Part A:
The engine works with two sources of heat, one hot (Qh) at a hot temperature (Th) and another cold (Qc) at a cold temperature (Tc). It is necessary so, the hot source will give energy to the fluid of the engine, and the cold source will be the source where these heat will dissipate and the fluid will return to its original temperature. So,
Qh > Qc, and Th > Tc
Part B:
The ideal heat engine is the one that can use the most amount of heat to transform it at work. It is characterized by Qh/Qc = Th/Tc.
The work is the useful energy, so it is the total heat (Qh) less the heat dissipated (Qc):
W = Qh - Qc
Part C:
The effiency is the useful energy divided by the total energy. Because W = Qh - Qc:
e = W/Qh
e = (Qh - Qc)/Qh
e = Qh/Qh - Qc/Qh
e = 1 - Qc/Qh