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Zina [86]
3 years ago
10

Which atom(s) is/are electrophilic in methanol (ch3oh)?

Chemistry
1 answer:
Zarrin [17]3 years ago
4 0

Let us understand the two terms

a) Nucleophile: which can give electrons it can be either negative or neutral

b) electrophile: which can accept electrons due to empty orbitals or incomplete octet. It can be positive or neutral

In methanol

Oxygen carries two lone pairs of electrons so it will be nucleophilic

while carbon and hydrogen being bonded to highly electronegative atom oxygen will be electrophilic

so answer is


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Alexxandr [17]
Either NaI or INa are acceptable formulae for sodium iodide, which is indeed a binary ionic compound. So, the correct answer would be B.
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Why is it that when you have a strong acid and weak base your equivalence point is less than 7
yKpoI14uk [10]

Explanation:

When a strong acid, say HCl reacts which a weak base, say NH_3, the reaction is shown below as:-

\text{NH}_3 (\text{aq}) + \text{HCl} (\text{aq}) \rightarrow {\text{NH}_4^+}(\text{aq}) + \text{Cl}^-(\text{aq})

The salt further reacts with water as shown below:-

\text{NH}_4^+ + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{NH}_3

Formation of \text{H}_3\text{O}^+ lowers the pH value of the solution as more hydrogen ions leads to less pH.

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3 years ago
Your answer should have the same number of significant figures is the start of measurement 400.08mL —— L
Eddi Din [679]
The answer would be .40008L

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1,000 mL in 1 L. 

As I understand, you are trying to convert 400.08mL to L. 

So this is how you are going to get it:

400.08mL x \frac{1L}{1,000mL}
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3 0
4 years ago
Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl
tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

  • Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

4 0
4 years ago
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tester [92]
<span>B.)the rate of reaction</span>
7 0
4 years ago
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