Answer:
29.4855 grams of chlorophyll
Explanation:
From Raoult's law
Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 457.45 mmHg ÷ 463.57 mmHg = 0.987
Mass of solvent (diethyl ether) = 187.4 g
MW of diethyl ether (C2H5OC2H5) = 74 g/mol
Number of moles of solvent = mass/MW = 187.4/74 = 2.532 mol
Let the moles of solute (chlorophyll) be y
Total moles of solution = moles of solute + moles of solvent = (y + 2.532) mol
Mole fraction of solvent = moles of solvent/total moles of solution
0.987 = 2.532/(y + 2.532)
y + 2.532 = 2.532/0.987
y + 2.532 = 2.565
y = 2.565 - 2.532 = 0.033
Moles of solute (chlorophyll) = 0.033 mol
Mass of chlorophyll = moles of chlorophyll × MW = 0.033 × 893.5 = 29.4855 grams
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
The displacement of the train after 2.23 seconds is 25.4 m.
<h3>
Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
Learn more about displacement here: brainly.com/question/2109763
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Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00

At x = 1.00 m

= 4J
Kinetic energy = (1/2)mv²

= 18J
Total energy will be =
4J + 18J = 22J
At x = 5

= -0.24J
Kinetic energy =

= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2

= 3.33 m/s
b) magnitude of force when x = 5.0m



At x = 5.0 m


= 0.016N
Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = 
indicates that the field is E for r = 2m
E =
(1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C