Answer:
3.1 * 10^-14
Explanation:
Note that E°cell = 0.0592/n log K
We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen
Anode reaction
H2(g) ----> 2H+ + 2e
Cathode reaction
Cd^2+(aq) + 2e -----> Cd(s)
E°cell = E°cathode - E°anode
E°cathode = –0.40 V
E°anode = 0 V
E°cell = –0.40 V - 0 V
E°cell = –0.40 V
E°cell = 0.0592/n log K
Where n=2 electrons transferred
–0.40 = 0.0592/2 log K
–0.40 = 0.0296 log K
log K = –0.40/0.0296
log K = -13.5135
K = Antilog ( -13.5135)
K = 3.1 * 10^-14
Given that the equipment is stationary, we should remove particles (such as food and soil) from around the equipment.
This step ensures that large particles are removed and do not hinder the remaining process of cleaning. After doing so, the next step is to remove the removable parts of the machinery. After this has been done, the cleaning steps may be undertaken.
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.
