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3 years ago

Fill in the coefficients that will balance the following reaction: a0PbI4 + a1CrCl3 → a2PbCl4 + a3CrI3

2 answers:

5 0

PbI₄ + CrCl₃ --> PbCl₄ + CrI₃

There is 1 Pb and Cr atom on both sides.
There are 3 Cl atoms on the left side and 4 on the right side.
There are 4 I atoms on the left side and 3 on the right side.

PbI₄ + 4CrCl₃ --> 3PbCl₄ + CrI₃
Cl atoms are now balanced

3PbI₄ + 4CrCl₃ --> 3PbCl₄ + 4CrI₃
All of the atoms are now balanced.

3PbI₄ + 4CrCl₃ --> 3PbCl₄ + 4CrI₃

Natasha_Volkova [10]3 years ago

3 0

A0 = 3, a1 = 4, a2 = 3, a3 = 4.

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In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

kirill115 [55]

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A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

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