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Neko [114]
2 years ago
9

A. Consider four different samples: aqueous LiI, molten LiI, aqueous AgI, and molten AgI. Current run through each sample produc

es one of the following products at the cathode: solid lithium, solid silver, or hydrogen gas. Match each sample to its cathodic product.B. Consider four different samples: aqueous NaBr, molten NaBr, aqueous NaF, and molten NaF. Current run through each sample produces one of the following products at the anode: liquid bromine, fluorine gas, or oxygen gas. Match each sample to its anodic product.
Chemistry
1 answer:
Rudiy272 years ago
5 0

Answer:

Explanation:

At the cathode

In case of molten AgI

Silver  will be collected

In case of molten LiI

lithium will be collected

in case of aqueous LiI,

hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺

in case of aqueous AgI,

Silver will be obtained at cathode because reduction potential of silver is more than H⁺

At the Anode  

In case of molten NaBr  

Bromine   will be collected

In case of molten NaF

Fluorine  will be collected

in case of aqueous NaBr ,

Bromine  will be collected as reduction potential of Br⁻ is less than O⁻²

in case of aqueous NaF ,

oxygen will be obtained  because reduction potential of F⁻  is more than O⁻² .

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A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var
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55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:

<em>pV = nRT</em>

Solve for <em>p</em>: <em>p = nRT/V</em>

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<em>T</em> = (265 + 273.15) K = 538.15 K

<em>V</em> = 500.0 mL = 0.5000 L

∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

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2 years ago
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The bonds in the compound MgSO4 can be described as
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The reaction of benzene with (CH3)3CCH2Cl in the presence of anhydrous aluminum chloride produces principally which of these?
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See explanation and image attached

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Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
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Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

B. 0.04 mol

C. 0.997 mol

D. 0.139 mol

E. 0.0681 mol

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The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

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= 100007.775 Pa

Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

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Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

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