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kicyunya [14]
3 years ago
7

Why the standard enthalpy change of hydration of copper sulphate cannot be measured directly ?

Chemistry
1 answer:
Juliette [100K]3 years ago
5 0

It is difficult to measure the enthalpy change of hydration accurately in a direct way because the hydration process can't be controlled directly. Instead, anhydrous and hydrated copper(II) sulfate can be dissolved in water. Each 'route' produces a solution of hydrated copper(II) sulfate.

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Which is the most to least soluble: NH3, NBr3, N2​
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Answer: https://quizlet.com/368481219/chem-2052-1-flash-cards/

I hope this helps! if not contact me for more help

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3 years ago
Overall do you think photosynthesis is endothermic or exothermic
blondinia [14]

Answer:

Endothermic hope that helps

Explanation:

6 0
4 years ago
Carbon exists in nature as three isotopes. Given the atomic masses and relative abundance of the isotopes, what is the average a
Ivan

The most abundant carbon isotope is carbon-12.

The relative atomic mass of carbon is 12.011, which is extremely close to 12.0. This means that the masses C-13, and C-14 are practically negligible when contributing to the relative atomic mass of carbon.

the C-12 isotope makes up 98.9% of carbon atoms, C-13 makes up 1.1% of carbon atoms, and C-14 makes up just a trace of carbon atoms as they are found in nature.

6 0
3 years ago
A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/
Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

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4 0
3 years ago
Prepare 15 mg/dl working standard solution from a stock solution of 20 mg/dl. State the volume of diluent and dilution.
Anastasy [175]

Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.

Recalling the principle:

initial volume x initial molarity = final volume x final molarity

Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:

Applying the dilution principle:

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final volume = 200/15

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More on dilution principle can be found here: brainly.com/question/11493179

4 0
3 years ago
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