I’m pretty sure that the answer is life, they all have life
Answer:
1. The length of the wire is halved.
the resistance of a conductor is directly proportional to it's length (L) as R - L. Thus doubling it's length will double it's resistance, while halving it's length would halve the resistance. Also the resistance of a conductor is inversely proportional to it's cross sectional area.
2. The area of cross-section of the conductor in increased.
On increasing the area of cross-section, resistance decreases. Thus is because resistance is inversely proportional to area.
3. The temperature of the conductor is increased.
With increasing temperature, the resistance of the wire increases as collisions within the wire increases and "slow" the flow of current... Since conductors typically display an increased resistivity with temperature increase, they have a positive temperature coefficient.
Filtration, he used a mesh to filter out the stuff
<span>Answer:
A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en).
contains
0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en
so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains
by the formula Cu(en)2^2+
0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+
Kf for Cu(en)2^2+ is 1x10^20.
so
1 Cu+2 & 2 en --> Cu(en)2^2+
Kf = [Cu(en)2^2+] / [Cu+2] [en]^2
1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2
[Cu+2] = [0.000300] / (1x10^20) (3.24 e-6)
Cu+2 = 9.26 e-19 Molar
since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
The answer is D 800 degrees Celsius
