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3 years ago

How is graphite different from other allotropes of carbon?

Chemistry

1 answer:

vladimir1956 [14]3 years ago

8 0

B) The carbon atoms in graphite are arranged in widely spaced layers

Explanation:

Graphite is different from other allotropes of carbon because the carbon atoms in graphite are arranged in widely spaced layers. Graphite and diamond are the two main allotropes of carbon that are closely related.

Both allotropes forms under extreme temperature and pressure. In geology, they are categorized as derivatives of changed rocks.
Graphite is different from carbon because its atoms are arranged in widely spaced layers.
Diamond has a packed and cross-linked structure that makes it rigid and very hard.

Learn more:

diamond brainly.com/question/10515084

#learnwithBrainly

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The product of adding two molecules of an alcohol to an aldehyde in the presence of acid is a(n):______

Anna35 [415]

Answer:

Explanation:

Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term.

To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile and second the water produced with the acetal must be removed from the reaction.

3 0

4 years ago

Read 2 more answers

WS 3.5 more Lewis structures

vlabodo [156]

Kinda confused what worksheet your on

6 0

3 years ago

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

4 years ago

If the solvent front is 68 mm and the ring front of an unknown is 48mm from the original spot, what is the rf value?

Nat2105 [25]

Given:

Distance of solvent front = 68 mm

Distance of unknown = 48 mm

To determine:

The rf value

Explanation:

The retention factor or the rf value is given by the ratio of distance traveled by the unknown to the distance traveled by the solvent front

RF = distance by unknown/distance by solvent

RF = 48/68 = 0.706

Ans: the RF value is 0.706

3 0

3 years ago

For 2Na (s) + Cl2(g) - 2NaCI (s) ,how many grams of sodium Chloride will be made from 115 g of sodium metal when reacted with ex

Solnce55 [7]

Given:

Mass of Na = 115 g

Excess Cl2

To determine:

Mass of NaCl produced

Explanation:

Given reaction is-

2Na(s) + Cl2(g) → 2NaCl(s)

Since Cl2 is in excess, Na will be the limiting reagent

As per the reaction stoichiometry Na:NaCl = 1:1

i.e. moles of Na reacted = moles of NaCl formed

Now, # moles of Na = mass of Na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

Therefore, moles of NaCl = 5

Molar mass of NaCl = 58 g/mol

Mass of NaCl = 5 moles * 58 g.mol-1 = 290 g

Ans: Amount of Nacl produced = 290 g

5 0

3 years ago