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Y_Kistochka [10]
3 years ago
6

Which enthalpy change (î"h) is correct for a substance changing from the liquid phase to the solid phase?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
7 0
I believe this should be the enthalpy of fusion. Enthalpy change is the amount of heat evolved or absorbed in a reaction. Enthalpy of fusion also called the heat of fusion resulting from providing heat energy or release of energy to a specific amount of a substance to change from solid state to liquid, without change in temperature.
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A brass doorknob has a mass of 80grams and a volume of 10cubic centimeters. What is its density?
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According to the Brinsted -Lowry. what is an acid ? what is a base ? ​
sasho [114]

Answer:

According to the Brønsted definition, an acid is a substance capable of donating a proton, and a base is a substance capable of accepting a proton. ... The species giving up the proton is HCl, an acid. The species accepting the proton is water, the base. The species Cl- is the conjugate base of HCl.

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5 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
3 years ago
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