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2 years ago

Will give brainliest

Chemistry

1 answer:

Dafna1 [17]2 years ago

3 0

sorry i know this:

4. 2 methyl 2 heptene

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Note.... the colliding molecules must possess certain minimum energy (threshold energy)to make the collision effective n..the ad

Oliga [24]

Answer:

<h3>I think, answer is threshold energy.</h3>

<h3>activation energy. </h3>

Explanation:

<h3>Hope it helps you....</h3>

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3 years ago

The bond dissociation energy to break 4 bond(s) in 1 mole of CH₄ molecules is:_____ **Any help would be greatly appreciated!**

frozen [14]

Answer:

The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ

Explanation:

Since there are 4 C-H bonds in CH₄, the bond dissociation energy of 1 mol of CH₄ is 4 × bond dissociation energy of one C-H bond.

From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ

So, the bond dissociation energy to break 4 bonds in 1 mol of CH₄ is 1644 kJ

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3 years ago

4AI+30 2 > 2AI2O3 What id the chemical reaction?

vesna_86 [32]

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3 years ago

Why can groundwater only collect in rock that is porous?

Galina-37 [17]

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2 years ago

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Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu

nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0_{cell}=E^0_{cathode}-E^0_{anode}$

$E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}$

$E^0_{cell}=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the standard Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole$

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0

3 years ago