Answer: Option (d) is the correct answer.
Explanation:
Electronegativity value of hydrogen is 2.2.
Electronegativity value of chlorine is 3.16.
Electronegativity value of carbon is 2.55.
Electronegativity value of oxygen is 3.44.
Electronegativity value of nitrogen is 3.04.
Electronegativity value of sodium is 0.93.
Electronegativity value of iodine is 2.66.
Therefore, calculate the electronegativity difference between the bonded atoms as follows.
- Electronegativity difference of HCl = Electronegativity value of chlorine - electronegativity value of hydrogen
= 3.16 - 2.2
= 0.96
- Electronegativity difference of CO = Electronegativity value of oxygen - electronegativity value of carbon
= 3.44 - 2.55
= 0.89
- Electronegativity difference of = Electronegativity value of nitrogen - electronegativity value of nitrogen
= 3.04 - 3.04
= 0
- Electronegativity difference of NaI = Electronegativity value of iodine - electronegativity value of sodium
= 2.66 - 0.93
= 1.73
So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.
Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.
Answer: an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.
Explanation:
Answer:
a) the minimun of acetic anhydride required for the reaction is 2.175 g (CH3CO)2O
b) V acetic anhydride = 2.010 mL
Explanation:
C6H4OHCOOH + (CH3CO)2O ↔ C9H8O4 + C2H4O2
⇒ mol salicylic acid = 2.94 g C6H4OHCOOH * ( mol C6H4OHCOOH / 138.121 g ) = 0.0213 mol C6H4OHCOOH
⇒ mol acetic anhydride = 0.0213 mol C6H4OHCOOH * ( mol (CH3CO)2O / mol C6H4OHCOOH ) = 0.0213 mol (CHECO)2O
⇒ g acetic anhydride = 0.0213 mol * ( 102.1 g/mol ) = 2.175 g CH3CO)2O
b) V = 2.175 g (CH3CO)2 * ( mL / 1.082 g ) = 2.010 mL (CH3CO)2O
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
Read more on equilibrium constant;
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