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3 years ago

How are the electron structures of boron (B) and aluminum (Al) similar?

2 answers:

4 0

The electron structures of boron and aluminum are similar because they share the same group, therefore they have the same amount of valence electron.

lbvjy [14]3 years ago

3 0

Answer:

- Both of their valence electrons are at p subshell.

- They have the first subshell full of electrons.

- Both of them have just 1 electron at the last p subshell.

Explanation:

Hello,

In this case, we could understand their electron structures by identifying their electron configurations as shown below:

$B^5\rightarrow 1s^2,2s^2,2p^1\\Al^{13}\rightarrow 1s^2,2s^2,2p^6,3s^2,3p^1$

In such a way, we could notice the following similarities:

- Both of their valence electrons are at p subshell.

- They have the first subshell full of electrons.

- Both of them have just 1 electron at the last p subshell.

Best regards.

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What is the molarity of the solution formed by dissolving 80. G of NAOH(s) into water to give a total volume of 4.00 l

Mrrafil [7]

Answer:

0.5 M

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 80 g

Volume of solution = 4 L

Molarity =?

Next, we shall determine the number of mole in 80 g of NaOH. This can be obtained as follow:

Mass of NaOH = 80 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 80 / 40

Mole of NaOH = 2 moles

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Mole of NaOH = 2 moles

Volume of solution = 4 L

Molarity =?

Molarity = mole / Volume

Molarity = 2/4

Molarity = 0.5 M

Therefore, the molarity of the solution is 0.5 M.

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3 years ago

Who discovered atomic structure

Yuliya22 [10]

John Dalton

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4 0

3 years ago

Which of the following is the conjugate acid of clo?

pshichka [43]

Explanation:

B. HCIO these is the answer

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3 years ago

Which criteria determine whether a heterogeneous mixture is a colloid or a suspension

Artemon [7]

whether the particles do not settle for an extended period of time

6 0

3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago