Answer:
when she squeezed oranges and the juice came out
Explanation:
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
Answer:
facultative anaerobes
Explanation:
Facultative anaerobes -
It is an organism , which is capable to prepare ATP via aerobic respiration , in the presence of oxygen , and can start the process of fermentation in the absence of oxygen , is the facultative anaerobes .
In the famous work by Louis pasteur , he was capable to ferment alcohol in the presence of facultative anaerobes .
Hence , from the given statement of the question,
The correct term is facultative anaerobes .
Answer:
C.) 76.1 grams
Explanation:
To find the mass of bromine, you need to (1) convert grams AlCl₃ to moles AlCl₃ (via molar mass), then (2) convert moles AlCl₃ to moles Br₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Br₂ to grams Br₂ (via molar mass). It is important to arrange your ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
Molar Mass (AlCl₃): 26.982 g/mol + 3(35.453 g/mol)
Molar Mass (AlCl₃): 133.332 g/mol
2 AlCl₃ + 3 Br₂ --> 2 AlBr₃ + 3 Cl₂
Molar Mass (Br₂): 2(79.904 g/mol)
Molar Mass (Br₂): 159.808 g/mol
42.3 g AlCl₃ 1 mole 3 moles Br₂ 159.808 g
------------------ x ----------------- x ---------------------- x ------------------- =
133.332 g 2 moles AlCl₃ 1 mole
= 76.0 g Br₂
*Our answers are slightly different most likely because we used slightly different molar masses*
