Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:
or,
First we have to calculate the equivalent weight of solute.
Molar mass of solute 
= 94.97 g/mole
Now we have to calculate the normality of solution.
Therefore, the normality of the solution is, 30.006 N
The molarity of solution made by diluting 26.5ml of 6.0ml hno3 to a volume of 250ml is calculated using the following formula
M1V1 = M2V2, where
M1 = molality 1 (6.00m)
V1= volume 1 (26.5 ml)
M2 = molarity 2(?)
v2=volume 2 (250)
M2 = M1V1/V2
M2= 6 x26.5/250 = 0.636 M
<span>The higher the molar mass is of the gas, the greater the density.
Cl2 is the answer</span>
Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
