Answer:
The atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.
Explanation:
As we travel along a period in a periodic table then the atomic radii decreases
This is because as we travel along a period we have that the atomic number of the atoms increases which means the the number of proton's increased
But the electron's add to the same outer shell throughout the period , which means the effective nuclear charge increases which pulls the outer electrons toward's the nucleus and the size decreases.
Therefore the atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.
A. Nonmetallic
B. Nonmetallic
C. Metallic
D. Nonmetallic
E. Metallic
Answer:
E²⁺
Explanation:
The group two contain alkaline earth metals.
There are six elements in group 2A.
Beryllium, Magnesium, calcium, strontium, barium and radium.
All members have two valance electrons.
They lose two valance electrons to complete the octet.
When they lose the two valance electrons they form cation X²⁺.
They react with halogens and form salt such as
MgCl₂, CaCl₂ etc.
Mg²⁺ Cl²⁻₂
The oxidation state of halogens are -1, while the elements of group two A shows +2 that's why two atoms of halogen are combine with one atom of alkaline earth metals and make the compound overall neutral.
All the alkaline earth metals have similar properties.
The energy can be shown as:
Q = ms dT
Whereas, m is the mass of block
s is specific heat
dT is change in temperature.
Copper block having the lowest specific heat and thus having the higher change in temperature and therefore having the higher final temperature.
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
