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Irina-Kira [14]
3 years ago
14

Element A has an electronegativity of 0.8 and element B has an electronegativity of 3.0. Which statement best describes the bond

ing in A3B? A) The AB bond is largely covalent with a on A. B) The AB bond is largely covalent with a δ+ on A. C) The compound is largely ionic with A as the cation. D) The compound is largely ionic with A as the anion.
Chemistry
1 answer:
Advocard [28]3 years ago
8 0

Answer:

C) The compound is largely ionic with A as the cation.

Explanation:

Pulings proposed the method to determine if the compound is ionic in nature or covalent in nature , by finding the difference between the electronegativity of the respective cation and anion .

The ion with higher electronegativity is the anion and the ion with lower electronegativity is the cation.

The electronegativity difference above 1.7 make the compound ionic in nature.

Hence, from the question ,

A is the cation and B is the anion.

And the electronegativity difference above 1.7 so the compound is ionic in nature.

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Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

PV=nRT

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between  pressure and number of moles of gas will be:

\frac{P_1}{P_2}=\frac{n_1}{n_2}

where,

P_1 = initial pressure of gas = 24.5 atm

P_2 = final pressure of gas = 5.30 atm

n_1 = initial number of moles of gas = 1.40 moles

n_2 = final number of moles of gas = ?

Now put all the given values in the above expression, we get:

\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}

n_2=0.301mol

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

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2 years ago
In a typical analysis, 15 ml of an aqueous solution containing an unknown amount of acetylcholine had a ph of 7.65. When incubat
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pH of the acetyl choline solution before incubation = 7.65

[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M

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The difference in concentration of hydronium ion before and after incubation

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This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.

Therefore the moles of acetylcholine = 15 mL * \frac{1L}{1000mL}*\frac{1.126*10^{-7}mol }{L}=1.689*10^{-9}mol


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