Answer:
Motile bacteria have flagella, while nonmotile bacteria do not.
Explanation:
Answer:
0.21 M. (2 sig. fig.)
Explanation:
The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.
How many moles of NaOH in the original solution?
,
where
is the number of moles of the solute in the solution.
is the concentration of the solution.
for the initial solution.
is the volume of the solution. For the initial solution,
for the initial solution.
.
What's the concentration of the diluted solution?
.
is the number of solute in the solution. Diluting the solution does not influence the value of
.
for the diluted solution.- Volume of the diluted solution:
.
Concentration of the diluted solution:
.
The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.
Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V
Calcium Chloride because it is a type 1 so the the anion ends with -ide
Answer:
nahhhhhhhhhhhhhhhhhhhhhhh xD
have a good day :)
Explanation: