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PolarNik [594]
3 years ago
8

Banced equation for AlBr3+Cl2=AlCl3+Br2

Chemistry
1 answer:
WITCHER [35]3 years ago
3 0

The balanced equation is 2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂

<em>Step 1</em>. Start with the <em>most complicated-looking formula</em> (AlBr₃?) and balance its atoms.

<em>Al</em>: Already balanced — 1 atom each side.

<em>Br</em>: We have 3 Br on the left-hand side and 2 Br on the right. the lowest common multiple of 3 and 2 is 6. Put a 2 in front of AlBr₃ and a 3 in front of Br₂.

2AlBr₃ + Cl₂ ⟶ AlCl₃ + 3Br₂

<em>Step 3</em>: Balance <em>Al</em>

We now need 2Al on the right to balance the 2 Al on the left. Put a 2 in front of AlCl₃.

2AlBr₃ + Cl₂ ⟶ 2AlCl₃ + 3Br₂

<em>Step 4</em>. Balance <em>Cl. </em>

We have 6 Cl on the right, so we need 6 Cl on the left. Put a 3 in front of Cl₂.

2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂

The balanced equation is

2AlBr₃ + 3Cl₂ ⟶ 2AlCl₃ + 3Br₂

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Answer: m = 50 g ZnSO4

Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.

0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

= 0.311 moles ZnSO4

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When V increases, p must decrease for k to remain constant.

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Convert .5km to inches
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1. A scientific hypothesis can become a theory if ______.
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When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
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Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
4 years ago
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