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1 year ago

What is the coordination number for atoms that pack within the diamond cubic crystal structure?.

1 answer:

6 0

Coordination numbers are defined for the crystalline structures that tell about the numbers of atoms bonded to the central atom. The coordination number of a diamond is 4.

<h3>What is a coordination number?</h3>

A coordination number has been defining the number of the ions, atoms, and molecules that are bonded to the central atom of the crystalline compound or a complex.

The diamond has 8 carbon atoms in its one unit and its coordination number is 4 as its lattice type is face-centered cubic that has carbons at the corners and face center of the structure. One carbon contributes 1/8 hence and the faces center contributes 1/2.

Therefore, the diamond has a coordination number of 4.

Learn more about coordination numbers, here:

brainly.com/question/27289242

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3 years ago

State two types of plastics and their uses.

Svetllana [295]

Thermoplastic and thermosetting

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1 year ago

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Which consequence of new fishing methods had the most immediate impact on the environment?

maxonik [38]

I believe the answer is D

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3 years ago

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Hello! Id like if someone could tell me this answer, thank you so much!

DaniilM [7]

Answer:

Option D. ZnCl₂ and H₂

Explanation:

From the question given above, the following equation was obtained:

2HCl + Zn —> ZnCl₂ + H₂

Products =?

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Reactants —> Products

Now, considering the equation from the question i.e

2HCl + Zn —> ZnCl₂ + H₂

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Thus, option D gives the correct answer to the question.

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2 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

2 years ago