Transition metals
Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one cation with different ionic charges. As an example, iron commonly forms two different ions
Answer:
NaCl>MgCl2> MgS>KBr
Explanation:
The smaller the cation, the higher the lattice energy of the compound
Answer: The balanced equation for the given reaction is
.
Explanation:
A chemical equation which contains same number of atoms on both reactant and product side.
For example, 
Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply
by 2 on reactant side and multiply
by 2. Hence, the equation will be re-written as follows.

Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.
Thus, we can conclude that the balanced equation for the given reaction is
.
That both of them are warm....at a certain temputurature causing one to fell,cozy
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.