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3 years ago

Explain why water and sodium oxide will have very different properties

Chemistry

1 answer:

Softa [21]3 years ago

4 0

Water and sodium oxide have different properties because of their nature as explained below.

<h3><u>Explanation:</u></h3>

Sodium oxide is a oxide of metallic sodium, while water is an oxide of hydrogen. So sodium oxide is a metallic oxide, while water is a non metallic oxide. Sodium oxide is a basic oxide, while water is neutral. As state of matter is concerned, sodium oxide is solid in normal room temperature, while water is liquid in normal room temperature. Water is a polar covalent molecule with partial charges on oxygen, but sodium oxide is an ionic molecule.

So all these factors contribute to very different properties of both sodium oxide and water.

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Rust forming on a piece of iron is an example of which of the following? *

kvasek [131]

Rust is an iron oxide and formed by the reaction of iron and oxygen in the presence of moisture. So the answer would be C

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3 years ago

HELP ASAS 15 POINTS

kramer

Answer:

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2 years ago

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

tia_tia [17]

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

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

The half-cell reactions are:

Oxidation half reaction (anode): $Zn\rightarrow Zn^{2+}+2e^-$

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

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction = ?

$[Zn^{2+}]$ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

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

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

Therefore, the cell potential for this reaction is 0.50 V

5 0

3 years ago

List and explain the four most important factors that control the relative reactivity of a carbonyl-containing functional group

ioda

Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.

Discussion:

1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.

2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.

3. Leaving group: Tetrahedral adduct fragmentation is encouraged by a better LG.

4. Steric effects: The nucleophilic attack on carbonyl carbon is delayed when sterically impeded.

Learn more about carbonyl here:

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8 0

1 year ago

Very urgent/easy chem question easy points

aev [14]

Sodium chloride is the answer

8 0

3 years ago

Read 2 more answers