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kifflom [539]
3 years ago
14

Vinegar has a density of 1.01 g/mL and freezes at approximately 2°C (36°F). It has a sour taste and is often used to make pickle

s. It can also be combined with baking soda to produce carbon dioxide gas. Which of the following describes a chemical property of vinegar? A. Vinegar has a sour taste. B. Vinegar freezes at 2°C. C. Vinegar can react with baking soda. D. Vinegar has a density of 1.01 g/mL.
Chemistry
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:

A vinegar can react with baking soda.

Explanation:

Chemical change:

The changes, that occur due to change in the composition of a substance and result in a different compound is known as chemical change.

These changes are irreversible

These changes occur due to chemical reactions

These may not be observed with naked eye

Example:

Vinegar and baking soda combine to produce CO₂

Combustion of fuel or wood: that oil or wood convert into energy, CO₂ and ash in case of wood

Boiling of egg: that change the chemical composition of protein in the egg

The reaction of Hydrogen and oxygen:

H₂ (g)  + O₂  (g) -------------------------------------> 2H₂O  (l)

Physical Change                                                                                              

The changes that occur only due to change in shape or form but their chemical or internal composition remain unchanged.

These changes were reversible.

They have same chemical property.

These changes can be observed with naked eye.

Example:

Density, taste, boiling points, freezing point, melting point are all physical properties.

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amount of solute dissolved at 30 C =

               =  23 / 100 * 39 = 8.97 g

Amount of extra solute dissolved = 8.97 - 4.29 = 4.7 g

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How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

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

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0
3 years ago
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