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3 years ago

What attractive forces must be overcome when CsI is dissolvedin liquid HF?

Chemistry

1 answer:

bagirrra123 [75]3 years ago

4 0

Answer:

The attractive forces must be overcome are :

Ion- ion interaction

Dipole - Dipole

Hydrogen Bonding

Explanation:

For the compound to dissolve the attractive forces existing between atoms of the compound must be reduced

CsI is ionic compound and its molecules are held together by ionic(electrostatic) force . These force must be weakened for its dissolution

Forces in HF :

1 .Hydrogen Bonding : In HF strong intermolecular Hydrogen Bonding exist between the electronegative F and Hydrogen

2. Dipole - dipole : HF is polar . So it is a permanent dipole and has dipole diople interaction

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What is a nonpolar covalent bond?

Vlad1618 [11]

Answer: The correct answer is Option D.

Explanation:

A non-polar covalent bond is defined as the bond which is formed between the atoms having no difference in electronegativity values. For Example: $Cl_2,H_2$ etc..

In this bond, the electrons are shared equally and $\Delta EN$ value is equal to 0.

Hence, the correct answer is Option D.

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2 years ago

Why is foam fight chemical or a physical change

nata0808 [166]

Answer:

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Explanation:

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2 years ago

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Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

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2 years ago

A pipe leaks water at a rate of 1.24 mL/s. What is the rate of the water leak in L/hr?

Umnica [9.8K]

Answer:

b)4.46 L/hr

Explanation:

To solve this question we need to convert the mL to liters (Using the conversion of 1000mL = 1L) and convert the time from seconds to hours (3600s = 1hr)

mL to L:

1.24mL/s * (1L / 1000mL) = 0.00124L/s

seconds to hours:

0.00124L/s * (3600s / 1hr) = 4.46L/hr

Right answer is:

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3 years ago

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2 years ago

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