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3 years ago

If a reaction mixture initially contains 0.195 m so2cl2, what is the equilibrium concentration of cl2 at 227 ???c?

1 answer:

6 0

Missing in your question the K constant value =2.99X10^-7
The reaction equation:
SOCl2 ⇄ SO2 +Cl2
initial 0.195 0 0
-X +X +X
final (0.195-X) X X

K= [SO2][Cl2]/SOCl2
by substitution
2.99X10^-7 = (X)(X)/(0.195-X)
2.99X10^-7 = X^2 / (0.195-X) by solving this equation
∴X = 0.00024 = 2.4 X10^-4
∴[Cl2] = 2.4X10^-4 m

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3 years ago

If 16.8 mL of the analyte H2C2O4 (oxalic acid) is titrated with 0.501 M KOH to 13.54 mL, determine the mass of the analyte.

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The balanced equation is:

$H_{2}C_{2}O_{4}+2KOH\rightarrow K_{2}C_{2})_{4}+H_{2}O$

The Molarity of Oxalic acid needed can be determined using: $M_{1}V_{1}=M_{2}V_{2}$

Here M1 = Molarity of Oxalic acid

V1 = Volume of oxalic acid

M2 = Molarity of KOH and V2 = Volume of KOH

M1 = ? , V1 = 16.8 mL , M2 = 0.501 M , V2 = 13.54 mL

Put the values and solve for M2,

$M_{1}V_{1}=M_{2}V_{2}$

$2\times M_{1}(16.8)=0.501\times 13.54$

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$M_{1}=\frac{0.501\times 13.54}{2\times 16.8}$

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0.202 M

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3 years ago

Consider the reaction when aqueous solutions of potassium hydroxide and ammonium nitrate are combined. The net ionic equation fo

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Answer: $OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

$KOH(aq)+NH_4NO_3(aq)\rightarrow KNO_3(aq)+NH_3(g)+H_2O(l)$

The equation can be written in terms of ions as:

$K^+(aq)+OH^-(aq)+NH_4^+(aq)+NO_3^-(aq)\rightarrow K^+(aq)+NO_3^-(aq)+NH_3(g)+H_2O(l)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The ions which are present on both the sides of the equation are potassium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is :

$OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)$

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4 years ago