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3 years ago

Determine the number of valence electrons for aluminum through the use of this table.

Chemistry

2 answers:

Zarrin [17]3 years ago

7 0

Answer:

Valence of Aluminium: +3

Explanation:

The only known number of valence electrons for the Al combined with other elements is +3. For example in substances such as Al2O3 (alumina).

julia-pushkina [17]3 years ago

3 0

There appears to be no table shown, but I can still answer. Aluminum has 3 valence electrons

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2 years ago

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What is the mole fraction of NaOH in an aqueous solution that

astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles = $\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of NaOH $(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water $(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

= $\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

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2 years ago

An electrochemical cell is powered by the half reactions shown below.

andrezito [222]

Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.

Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.

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3 years ago