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2 years ago

Besides their shape, how does a 2p orbital differ from a 3d orbital? chegg.

Chemistry

1 answer:

Blababa [14]2 years ago

5 0

Answer:

The 3p orbitals have the same general shape and are larger than 2p orbitals, but they differ in the number of nodes. You have probably noticed that the total number of nodes in an orbital is equal to n−1 , where n is the principal quantum number. Thus, a 2p orbital has 1 node, and a 3p orbital has 2 nodes.

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Turgor, an internal stimulus, refers to —

VladimirAG [237]

Answer:

which has a lower freeging point oxgyen or ethanol

6 0

2 years ago

Question 2 (Multiple Choice Worth 2 points)

faust18 [17]

Answer:

$\pi$ r² x h

Explanation:

A can is cylindrical in nature. Using the formula of the volume of the can, we can find this unknown volume.

The volume of cylinder is given as:

Volume of a cylinder = Area x height

Volume of a cylinder = $\pi$ r² x h

Therefore density of the can;

Density = $\frac{mass}{volume}$

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2 years ago

Explain why mineral oil which is a mixture of hydrocarbons does not dissolve in water

Mkey [24]

Oil molecules like to stick to other oil molecules more than they like to stick to water molecules. This is because mineral oil has nonpolar molecules and water has polar ones.

3 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

5 0

3 years ago

Magnesium reacts with Oxygen gas forming Magnesium oxide as shown;

LuckyWell [14K]

Molar mass:

O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol

<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)

mass of O2 = 5.00 x 32 / 2 x 24.0

mass of O2 = 160 / 48

= 3.33 g of O2

hope this helps!

5 0

3 years ago