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2 years ago

11

100 mL of a strong acid is completely neutralized by 100 mL of a strong base. The observed products are salt and water. Upon fur

ther investigation, the solution has a pH > 7. How is this possible?

1 answer:

ser-zykov [4K]2 years ago

3 0

Answer: The concentration of base is higher than the concentration of the acid.

Explanation:

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6-methylhept-1-en-3-one

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2 years ago

Which of the following is NOT an example of potential energy?

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A. car moving down the highway

Explanation:

Potential energy is when the energy is being held in, as in, it is very still, and not moving.

When a car is moving down a highway, it is releasing kinetic energy, instead of potential.

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2 years ago

Why should you prefer dry ashing to wet ashing of food sample such as cabbage

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2 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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3 years ago