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Mumz [18]
3 years ago
11

An unknown element, X, reacts with sodium to form the compound Na2X. In other compounds, this element also can accommodate up to

12 electrons rather than the usual octet. What element could X be?
A. Sr
B. O
C. Cl
D. Po
Chemistry
1 answer:
Andreas93 [3]3 years ago
7 0

Answer:

B. O

Explanation:

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The attraction which exists between carbon diozide molecules in solid carbon dioxide is due to
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Van der waal's  force of attraction

Explanation:

Van der waal's  force of attraction

Forces of Van der Waals involve attraction and repulsion among atoms, particles, and surfaces as well as many other intermolecular forces. These vary from covalent and ionic bonding in that they are caused by similarities in surrounding particle fluctuating polarizations (a result of quantum dynamics).

8 0
3 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

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