Answer:
D. Quantitative
Explanation:
Quantitative observation deals with numbers, or amounts.
Qualitative observation deals with de- scriptions that cannot be expressed in numbers.
The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:

=

Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).

=

∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
⇒ P₂ = (4.12 kPa · °C) (50 °C)
=
206 kPa
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is
206 kPa
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.
Liquid petrol does not ignite spontaneously when exposed to the air because there is not enough molecules that contains energy to reach the activation energy. As a result, the reaction cannot proceed and a supply of energy should be added for the reaction to occur.
Answer:
a it rarely reach with other elements