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2 years ago

Which of the following statements about alkenes is not correct?

Chemistry

1 answer:

professor190 [17]2 years ago

8 0

Option-C: They react mainly by substitution.

Explanation:
Alkene doesn't give substitution reactions because they are non polar in nature. The double bond in alkene is responsible for Electrophillic Addition reactions as it electron rich and nucleophilic in nature. Reaction of Alkene is given below,

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A newly discovered unicellular organism from acidic mine drainage is found to contain a cell wall, a plasma membrane, two flagel

makvit [3.9K]

Answer:

The correct option is D) a motile eukaryote

Explanation:

Peroxisomes are structures which are present only in the plant or animal cells. Hence, based on this observation we can easily tell that the organism which is newly discovered is a eukaryote. Both plants and animals are eukaryotes.

Flagella are structures which are thin and hair shaped. Flagella allow movement of a cell, mostly they allow a whip-like movement. Based on this observation we will conclude that the organism discovered is a motile eukaryote.

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2 years ago

Which aqueous solution will cause the bulb to light?

likoan [24]

LiOH(aq)

Hope this helps :)

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2 years ago

For the following reaction, 53.7 grams of iron(III) oxide are allowed to react with 22.8 grams of aluminum. iron(III) oxide (s)

ZanzabumX [31]

Answer:

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Iron (III) oxide is a limiting reagent i.e $Fe_2O_3$ .

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

Explanation:

iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (s)

$Fe2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)$

Moles of iron(III) oxide : $\frac{53.7 g}{160 g/mol}=0.3356 mol$

Moles of aluminium : $\frac{22.8 g}{27 g/mol}=0.8444 mol$

According to recation, 1 mole of iron(III) oxide reacts with 2 moles of aluminum.

Then 0.3356 moles of iron(III) oxide will react with:

$\frac{2}{1}\times 0.3356 mol=0.6712 mol$ of aluminum.

As we can see that moles of iron(III) are in limiting amount.Hence iron(III) oxide is a limiting reagent i.e $Fe_2O_3$ and aluminum in the an excessive reagent.

Amount of aluminum oxide will depend upon moles of limiting reagent that is iron(III) oxide.

According to reaction , 1 mole iron(III) oxide gives 1 moles of aluminum oxide.

Then 0.3356 moles will give:

$\frac{1}{1}\times 0.3356 mol=0.3356 mol$ of aluminum oxide

Mass of 0.3356 moles of aluminum oxide:

0.3356 mol × 102 g/mol = 34.23 g

34.23 grams is the maximum amount of aluminum oxide that can be formed.

Moles of excessive reagent left = 0.8444 mol - 0.6712 mol = 0.1732 mol

Mass of 0.1732 moles of aluminum :

0.1732 mol × 27 g/mol = 4.6764 g

4.6764 grams is the amount of the aluminum which remains after the reaction is complete

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2 years ago

Scientists are often interested in knowing the molar heat of combustion the heat released during the combustion of one molar of

Butoxors [25]

Answer:

.004993

Explanation:

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2 years ago

Read 2 more answers

Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method.

tatiyna

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

(a) $2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O$

(b) $2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O$

(c) $2Cr^{3+}+7H_2O+3Cl_2\rightarrow Cr_2O_7^{2-}+14H^++6Cl^-$

(d) $2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is,

$O_2(g)+Pb(s)\rightarrow H_2O(l)+Pb^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2\rightarrow H_2O$

Now balance oxygen atom on both side.

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2\rightarrow 2H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Pb\rightarrow Pb^{2+}$

Reduction : $O_2+4H^+\rightarrow 2H_2O$

Now balance the charge.

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

Reduction : $O_2+4H^++4e^-\rightarrow 2H_2O$

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

Oxidation : $2Pb\rightarrow 2Pb^{2+}+4e^-$

Reduction : $O_2+4H^++4e^-\rightarrow 2H_2O$

The balanced chemical equation will be,

$2Pb+O_2+4H^+\rightarrow 2Pb^{2+}+2H_2O$

(b) The given chemical reaction is,

$NO_3^-(aq)+Sn(s)\rightarrow NO(g)+Sn^{2+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-\rightarrow NO$

Now balance oxygen atom on both side.

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-\rightarrow NO+2H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn\rightarrow Sn^{2+}$

Reduction : $NO_3^-+4H^+\rightarrow NO+2H_2O$

Now balance the charge.

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

Reduction : $NO_3^-+4H^++4e^-\rightarrow NO+2H_2O$

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

Oxidation : $2Sn\rightarrow 2Sn^{2+}+4e^-$

Reduction : $NO_3^-+4H^++4e^-\rightarrow NO+2H_2O$

The balanced chemical equation will be,

$2Sn+NO_3^-+4H^+\rightarrow 2Sn^{2+}+NO+2H_2O$

(c) The given chemical reaction is,

$Cl_2(g)+Cr^{3+}(aq)\rightarrow Cl^-(aq)+Cr_2O_7^{2-}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $2Cr^{3+}\rightarrow Cr_2O_7^{2-}$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance oxygen atom on both side.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance hydrogen atom on both side.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^+$

Reduction : $Cl_2\rightarrow 2Cl^-$

Now balance the charge.

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^-$

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

Oxidation : $2Cr^{3+}+7H_2O\rightarrow Cr_2O_7^{2-}+14H^++6e^-$

Reduction : $3Cl_2+6e^-\rightarrow 6Cl^-$

The balanced chemical equation will be,

$2Cr^{3+}+7H_2O+3Cl_2\rightarrow Cr_2O_7^{2-}+14H^++6Cl^-$

(d) The given chemical reaction is,

$F_2(g)+Mn^{2+}(aq)\rightarrow F^-(aq)+MnO_4^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Mn^{2+}\rightarrow MnO_4^-$

Reduction : $F_2\rightarrow 2F^-$

Now balance oxygen atom on both side.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-$

Reduction : $F_2\rightarrow 2F^-$

Now balance hydrogen atom on both side.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+$

Reduction : $F_2\rightarrow 2F^-$

Now balance the charge.

Oxidation : $Mn^{2+}+4H_2O\rightarrow MnO_4^-+8H^+5e^-$

Reduction : $F_2+2e^-\rightarrow 2F^-$

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $2Mn^{2+}+8H_2O\rightarrow 2MnO_4^-+16H^+10e^-$

Reduction : $5F_2+10e^-\rightarrow 10F^-$

The balanced chemical equation will be,

$2Mn^{2+}+8H_2O+5F_2\rightarrow 2MnO_4^-+16H^++10F^-$

5 0

2 years ago