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3 years ago

15 POINTS FOR ANYONE WHO CAN ANSWER THIS!!!!!! plZ

Chemistry

1 answer:

yaroslaw [1]3 years ago

8 0

Answer:

A. 2Al(s) + 3Cl2(g) —> 2AlCl3(s)

B. 2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

C. H2(g) + Br2(l) —> 2HBr(g)

Explanation:

A. Reaction between aluminium metal and chlorine gas.

Al(s) + Cl2(g) —> AlCl3(s)

The above equation can be balance as follow:

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of AlCl3 as shown below:

Al(s) + 3Cl2(g) —> 2AlCl3(s)

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al(s) + 3Cl2(g) —> 2AlCl3(s)

Now the equation is balanced.

B. Reaction between lithium metal and liquid water.

Li(s) + H2O(l) —> LiOH(aq) + H2(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and a total of 3 atoms on the right side. It can be balance by putting 2 in front of H2O and 2 in front of LiOH as shown below:

Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

There are 2 atoms of Li on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Li as shown below:

2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

Now, the equation is balanced.

C. Reaction between gaseous hydrogen and liquid bromine.

H2(g) + Br2(l) —> HBr(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and 1 atom on the right. It can be balance by putting 2 in front of HBr as shown below:

H2(g) + Br2(l) —> 2HBr(g)

Now the equation is balanced.

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Balance each of the following in BASE: Show work

kupik [55]

Answer : The balanced chemical equation in a basic solution are,

(A) $2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) $2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) $4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) $Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

Explanation :

(A) The given chemical reaction is :

$CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)$

The oxidation-reduction half reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}$

or,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) The given chemical reaction is :

$NO_2(g)\rightarrow NO_3^-(aq)+NO_2^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $NO_2+2OH^-\rightarrow NO_3^-+H_2O+1e^-$

Reduction : $NO_2+1e^-\rightarrow NO_2^-$

The electrons in both the reactions are balanced. Now added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) The given chemical reaction is :

$Zn(s)+NO_3^-(aq)\rightarrow Zn(OH)_4^{2-}(aq)+NH_3(g)$

The oxidation-reduction half reaction will be :

Oxidation : $Zn+4OH^-\rightarrow [Zn(OH)_4]^{2-}+2e^-$

Reduction : $NO_3^-+6H_2O+8e^-\rightarrow NH_3+9OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 4 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) The given chemical reaction is :

$Br_2(l)\rightarrow BrO_3^-(aq)+Br^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Br_2+12OH^-\rightarrow 2BrO_3^-+6H_2O+10e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^-$

In order to balance the electrons, we multiply the reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

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3 years ago

What is the name of the compound cs (mno4)2

N76 [4]

D. cadmium permangantie

3 0

3 years ago

Please answer! need it soon

mario62 [17]

Answer to what maybe I can help you

6 0

3 years ago

What is the henry's law constant for co2 at 20∘c? express your answer to three significant figures and include the appropriate u

Gennadij [26K]

Actually, Henry's Law is an empirical value. It means that it was not obtained out of raw calculations or correlations. This was gathered from experimental results. Hence, you can search its data. At standard temperature of 25°C (298 K),

k = k°e^[2400(1/T - 1/T°)], where k° = 29.4 L·atm/mol
Substituting the values so that T would be in 20°C or 293 K,
k = (29.4 L·atm/mol)e^[2400(1/293 - 1/298)]
k = 33.7 L·atm/mol

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3 years ago

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Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.

jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

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3 years ago