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Elina [12.6K]
3 years ago
13

Label each bond in the following compounds as ionic or covalent (10) (a) F2 (b) LiBr (c) NaNH2 (d) CH3CH3

Chemistry
1 answer:
Zigmanuir [339]3 years ago
5 0

Explanation:

Covalent bond -

This type of bonding takes place between the atoms of same or similar electronegativity , is referred to as covalent bonding .

Ionic bonding -

This type of bonding takes place between the atoms of different electronegativity , is referred to as Ionic bonding .

From the question ,

( a ) F₂ - Covalent bonding ( same atom , hence , same the electronegativity )

( b ) LiBr - Ionic bonding ( the electronegativity difference between Li and Br is high )

( c ) NaNH₂ - Ionic bonding ( the electronegativity difference between sodium and Nitrogen is high  )

( d )CH₃CH₃ - Covalent bonding  ( since , the electronegativity difference between carbon and hydrogen is similar )

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3 years ago
The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

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5 0
3 years ago
A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25°.
ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

pOH=-\log[OH^-]

1.00=-\log[OH^-]

[OH^-]=10^{-1.00} M=0.100 M

KOH(aq)\rightarrow K^+(aq)+OH^-(aq)

[KOH]=[OH^-]=[K^+]=0.100 M

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

Molarity=\frac{Moles}{Volume(L)}

0.100 M=\frac{n}{0.800 L}

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

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