Answer:
-3.617 °C
Explanation:
Step 1: Given data
Mass of water (m): 210.0 g
Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)
Specific heat of water (c): 4.184 J/g.°C
Temperature change (ΔT): ?
Step 2: Calculate the temperature change
We will use the following expression.
Q = c × m × ΔT
-3178 J = 4.184 J/g.°C × 210.0 g × ΔT
ΔT = -3.617 °C
The ionic radius increases.
Each time you move down to the next Period, you are adding an extra shell of electrons.
Each shell is further from the nucleus than the one before it, so the .
The image below illustrates the increase in ionic radius as you go down a Group.
The molar solubility is 7.4×
M and the solubility is 7.4× g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→ 
+ 
S
- S S
Ksp = [ ] [ ] = [S] [ S ] = 
S = √ Ksp = √ 5. 5×
= 7.4×
The solubility =7.4× g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of is dissociates and form one mole of each and ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4× mol/L = 7.4× M
To learn more about molar solubility ,
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The Group number of a non-transition metal can be used to find the number of valence electrons in an atom of that element. The ones place of the group number is the number of valence electrons in an atom of these elements
