Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
<span>4) Formation of a gas
When methane burns, it produces the gases water vapor and carbon dioxide.
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Models help us to understand systems and their properties
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
fault-block mountains. hope this helps
