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ollegr [7]
3 years ago
7

As the temperature of a liquid increases, its viscosity _____. increases stays the same decreases

Chemistry
2 answers:
Olin [163]3 years ago
6 0

Answer: Option (c) is the correct answer.

Explanation:

Viscosity is defined as the ability of a liquid to resist its flow. When a substance has high viscosity then it is known as a viscous substance.

For example, honey has high viscosity.

But when we increase the temperature then there occurs more number of collisions between the molecules. Due to this there will be increase in kinetic energy of the molecules and they will be able to move the fluid easily.

As,    Kinetic energy = \frac{3}{2}kT

Thus, we can conclude that as the temperature of a liquid increases, its viscosity decreases.

Valentin [98]3 years ago
5 0
As the temperature of a liquid increases, its viscosity decreases.
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Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

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