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4 years ago

Predict which atom has a higher ionization energy. CL or I

Chemistry

2 answers:

maw [93]4 years ago

8 0

Answer:

The answer is chlorine (Cl)

Explanation:

Ionization energy is the minimum amount of energy required to remove a valence electron from an atom in its gaseous state. The closer the valence electron (negatively charged) is to the nucleus (positively charged), the higher the ionization energy required to remove it, hence ionization energy generally decreases down the group.

Between the two, chlorine comes first in the group before iodine and hence has a higher ionization energy (as explained earlier) than Iodine

bija089 [108]4 years ago

7 0

I'm pretty sure your answer is CL.

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Cuál es tejidos conforman el sistema de tejido fundamental

ELEN [110]

Explanation:

Los tejidos que conforman el tejido fundamental son los siguientes:

-Parénquima

-Colénquima

-Esclerénquima.

El tejido fundamental constituye la parte principal del cuerpo de la planta porque intervienen en funciones tan importantes para el desarrollo y sostenimiento de la planta como el proceso de la fotosíntesis, el almacenamiento de agua y de nutrientes .

5 0

3 years ago

If you add 14.22ml of 2.97m hcl (42.2mmol) to an antacid, then neutralize the excess acid with 5.00ml of 0.1055 m naoh (0.528mmo

Vinvika [58]

Given data:

Volume of HCl = 14.22 ml

Molarity of HCl = 2.97 M

mmoles of HCl = 14.22 * 2.97 = 42.2 mmoles

Volume of NaOH = 5.00 ml

Molarity of NaOH = 0.1055 M

mmoles of NaOH = 5.00 *.1055 = 0.5275 mmoles

Since HCl and NaOH combine in a 1:1 ratio

# moles of NaOH = # moles of excess HCl that is neutralized = 0.5275 moles

Now, the total moles of HCl taken = # mmoles HCl neutralized by antacid + # mmoles of excess HCl

42.2 = mmoles HCl neutralized by antacid + 0.5275

Therefore,

mmoles of HCl neutralized by antacid = 42.2 - 0.5275 = 41.6725 mmoles = 41.7 mmoles

5 0

3 years ago

When two molecules of methanol (CH3OH) react with oxygen, they combine with three O2 molecules to form two CO2 molecules and fou

Alex17521 [72]

Answer:

188

Explanation:

For every 2 molecules of methanol reacted, 4 molecules of water are formed. Use this relationship to solve.

2/4 = 94/x

2x = 376

x = 188

188 molecules of water will be formed.

3 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

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

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

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

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

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

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

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

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

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

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

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.

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

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

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$