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aleksandrvk [35]
2 years ago
9

How can increase the rate of the preparation copper i iodide?​

Chemistry
1 answer:
BARSIC [14]2 years ago
5 0

Answer:

Preparation. Copper(I) iodide can be prepared by heating iodine and copper in concentrated hydriodic acid, HI. In the laboratory however, copper(I) iodide is prepared by simply mixing an aqueous solution of potassium iodide and a soluble copper(II) salt such copper sulfate.

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Which is NOT a property of gold?
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Rareness, the others describe what the gold is and its properties.

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If I have a 50.0 liter container that holds 45 moles of gas at a
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How do I find the formula for Phosphorus Tetroxide
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The vapor pressure of pure pentane and pure hexane are 425 torr and 151 torr respectively. equal moles of pentane and hexane are
nadezda [96]

The mole fraction of pentane in the vapor is 0.291

Vapour pressure rises with temperature and is a measurement of a substance's propensity to transform into a gaseous or vapour state. The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapour pressure present at the liquid's surface.

The number of moles of a particular component in the solution divided by the overall number of moles in the sample solution is known as the mole fraction.

Using the formula for vapour pressure,

vapour pressure = P^{}°_{hex} × X_{hex} + P^{}°_{pen} × X_{pen}

vapour pressure = 151 ×  (1-X_{pen}) + 425 × X_{pen}

240 = 151 - 151X_{pen} + 425X_{pen}

240 - 151 = -  151X_{pen} + 425 X_{pen}

89 = 274 X_{pen}

\frac{89}{274} = X_{pen}

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Therefore,  the mole fraction of pentane in the vapor is 0.291

Learn more about vapour pressure here;

brainly.com/question/25699778

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1 year ago
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