Answer:
Francium (Fr)
Explanation:
From the given choices, francium will have the lowest ionization energy.
Ionization energy is the energy required to remove the most loosely held electron within an atom.
The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.
- Generally on the periodic table, ionization energy increases from left to right on the table
- As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
- Down the group, the value reduces.
- Since Francium is the most metallic of all the given choices, it has the highest ionization energy.
Answer:
C.) 35
Explanation:
The mass is made up of the total protons and neutrons in an atom. Protons and neutrons both have a mass of 1 amu. Electrons are not included in this measurement because they have an insignificant mass (practically 0).
(17 protons x 1 amu) + (18 neutrons x 1 amu) = 35 amu
Therefore, if an atom contains 17 protons and 18 neutrons, the mass should be 35 amu.
<u>Answer:</u> The correct answer is ![3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V](https://tex.z-dn.net/?f=3Ag%5E%2B%28aq.%29%2BCr%28s%29%5Crightarrow%203Ag%28s%29%2BCr%5E%7B3%2B%7D%28aq.%29%3BE%5Eo_%7Bcell%7D%3D%2B1.53V)
<u>Explanation:</u>
We are given:
![Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V](https://tex.z-dn.net/?f=Cr%5E%7B3%2B%7D%28aq.%29%2B3e%5E-%5Crightarrow%20Cr%28s%29%3BE%5Eo%3D-0.73V%5C%5C%5C%5CAg%5E%2B%28aq.%29%2Be%5E-%5Crightarrow%20Ag%28s%29%3BE%5Eo%3D%2B0.80V)
The substance having highest positive
potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.
Chromium will undergo oxidation reaction and will get oxidized.
The half reactions for the above cell is:
Oxidation half reaction: ![Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V](https://tex.z-dn.net/?f=Cr%28s%29%5Crightarrow%20Cr%5E%7B3%2B%7D%2B3e%5E-%3BE%5Eo_%7BCr%5E%7B3%2B%7D%2FCr%7D%3D-0.73V)
Reduction half reaction:
( × 3)
Net equation: ![3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)](https://tex.z-dn.net/?f=3Ag%5E%2B%28aq.%29%2BCr%28s%29%5Crightarrow%203Ag%28s%29%2BCr%5E%7B3%2B%7D%28aq.%29)
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:
![E^o_{cell}=E^o_{cathode}-E^o_{anode}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7Bcathode%7D-E%5Eo_%7Banode%7D)
Putting values in above equation, we get:
![E^o_{cell}=0.80-(-0.73)=1.53V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D0.80-%28-0.73%29%3D1.53V)
Hence, the correct answer is ![3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V](https://tex.z-dn.net/?f=3Ag%5E%2B%28aq.%29%2BCr%28s%29%5Crightarrow%203Ag%28s%29%2BCr%5E%7B3%2B%7D%28aq.%29%3BE%5Eo_%7Bcell%7D%3D%2B1.53V)
Answer:
volume of box is 38.81 cm³.
Explanation:
Given data:
Width of box = 4.5 cm
Height of box = 5.750 cm
Length of box = 1.50 cm
Solution:
Formula:
Volume = length × height × width
by putting values,
V = 1.50 cm × 5.750 cm× 4.5 cm
V = 38.81 cm³
Thus, the volume of box is 38.81 cm³.