<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>
Answer:
24.45 L
Explanation:
From General gas equation,
PV/T = P'V'/T'.................... Equation 1
Where P = initial pressure, V = Initial volume, T = Initial Temperature, P' = Final pressure, V' = Final Volume, T' = Final Temperature.
make V' the subject of the equation.
V' = PVT'/(TP')................. Equation 2
Given: P = 760 torr, V = 22.4 L, T = 0 °C = (0+273) = 273 K, T' = 25 °C = (25 + 273) = 298 K, P' = 760 torr
Substitute into equation 2
V' = 760(22.4)(298)/(760×273)
V' = 24.45 L.
Hence the volume of the gas = 24.45 L
Answer:
2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Explanation:
Given data:
Mass of sample = 142.1 g
Initial temperature = 25.5°C
Final temperature = 46°C
Specific heat capacity of Al = 0.90 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 46°C - 25.5°C
ΔT = 20.5°C
Q = 142.1 × 0.90 J/g.°C × 20.5°C
Q = 2621.75 j
Thus, 2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Answer:
Crest
Explanation:
The highest point on a wave is called the crest and the lowest point is called the trough.