Answer:
Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below
The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.
Explanation:
The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.
In the Left Beaker (Left half cell), their is less concentration
Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side
In the Right Beaker (right half cell), their is more concentration
Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side
Electrons will flow from Left to Right direction.
Yes the answer is correct the plant uses the energy to grow and fertilize other plants
Answer:
Explanation:
Explanation:
All you have to do here is use the ideal gas law equation, which looks like this
P
V
=
n
R
T
−−−−−−−−−−
Here
P
is the pressure of the gas
V
is the volume it occupies
n
is the number of moles of gas present in the sample
R
is the universal gas constant, equal to
0.0821
atm L
mol K
T
is the absolute temperature of the gas
Rearrange the equation to solve for
T
P
V
=
n
R
T
⇒
T
=
P
V
n
R
Before plugging in your values, make sure that the units given to you match those used in the expression of the universal gas constant.
In this case, the volume is given in liters and the pressure in atmospheres, so you're good to go.
Plug in your values to find
T
=
3.10
atm
⋅
64.51
L
9.69
moles
⋅
0.0821
atm
⋅
L
mol
⋅
K
T
=
251 K
−−−−−−−−−
The answer is rounded to three
Answer:
The answer is A. Lithium and Sodium
Answer : The reaction rate at
are 
Solution : Given,
Mass of tablet = 1000 mg
Volume of water = 0.200 L
Formula used :

Now we have to calculate the reaction rate at different temperatures and reaction time.




Therefore, the reaction rate at
are 