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soldi70 [24.7K]
3 years ago
9

Aqueous solutions of aluminum sulfate and barium chloride are mixed, resulting in the precipitate formation of barium sulfate wi

th aqueous aluminum chloride as the other product. (Use the lowest possible coefficients.
Chemistry
1 answer:
denpristay [2]3 years ago
3 0

Answer:

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

Explanation:

The question wants you to use the lowest possible coefficients to balance the equation.

According to the question the reaction is as follows ;

Generally, writing the chemical formula requires one to exchange the charge between the cations and anions involved. Example aluminum sulfate has Al3+ and (SO4)2- . cross multiply the charges to get Al2(SO4)3

aluminum sulfate → Al2(SO4)3

barium chloride → BaCl2

barium sulfate → BaSO4

aluminum chloride → AlCl3

Al2(SO4)3(aq) + BaCl2(aq) → BaSO4(s) + AlCl3(aq)

To balance a chemical equation one have to make sure the number of atom of element on the reactant side(left) is equal to the number of atom of elements on the product side(right).

Now, the equation can be be balance with the lowest coefficient as follows;

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

The bold numbers is the coefficient use to balance the equation.

The number of atom on the reactant side is equal to the product side. Using aluminium atom as a case study the number of aluminium atom on the reactant side is 2 and the on the product  side it is also  2.

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
3 years ago
Name the process that changes matter into one or more new substances
Brums [2.3K]
Chemical change or process
6 0
3 years ago
Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4
Anvisha [2.4K]

Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃      →     Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

                       Pb(SO₄)₂        :         Li₂SO₄

                            1                :             2

                          2.25           :          2/1×2.25 = 4.5 mol

                       LiNO₃            :             Li₂SO₄

                           4                :                2

                           9.62           :             2/4×9.62 = 4.81 mol

Pb(SO₄)₂  produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of  Li₂SO₄.      

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mash [69]

The answer is 3.592?

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A nickle is about 2 millimeters thick, or 2/1000 of a meter.how many nanometers is this?
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You know that you can just google it right? It has this converter.
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