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fgiga [73]
2 years ago
8

Diamond cannot conduct electricity but graphite can,why​

Chemistry
1 answer:
zavuch27 [327]2 years ago
8 0

Diamond does not conduct electricity because it has no delocalized electron to conduct electricity.

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m_a_m_a [10]

Answer:

B

Explanation:

living things are not required to move

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An isomer with a branched structure will require:
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The  branched  structure  isomer  will require   less  energy  to  melt  than the straight  chain  isomer

    explanation
Branched  structure  isomer    has  weak  intermolecular  forces  of attraction  as  compared  to  straight  chain  isomers.  In addition the  branched  isomer  has  a  low boiling point as  compared to  straight chain isomers. Since   boiling require  the of  the  intermolecular  forces tend to have  lower  boiling  point   than straight chain
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Why do some elements form double and triple bonds during bonding
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Explanation:

Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more  the reason is the  interaction between the p orbitals of the combining atoms. for example  A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.

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3 years ago
Draw the structure of 4-methylcycloheptanol.
Furkat [3]

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<em>Answer is in the attachment</em>

6 0
2 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0
3 years ago
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