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bogdanovich [222]
4 years ago
6

Our system is the liquid water contained in a bath tub. The drain is open at the bottom while water is being poured into it at a

n equivalent rate so that the water level within the bathtub is not rising. Is this process steady-state?
Chemistry
1 answer:
Ne4ueva [31]4 years ago
8 0

Answer:

The process described in the problem is steady-state.

Explanation:

The steady-state can be described by the conservation of mass. In a bathtub it will be the same as In=Out+ Accumulation. Since the water has already reached the adequate level and the drain is open, the accumulation is equal to zero, and therefore it is at steady state. This also means that In = Out which is also described in the problem: "the water is poured into the bathtub at an equivalent rate of the water being drained".

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WHICH IS WHICH?
Aleksandr-060686 [28]

Answer:

- First: element.

- Second: Compound.

- Third: element.

- Fourth: Compound.

Explanation:

Hello.

In this case, since elements are composed the same atom and compounds are composed by two different atoms, we can see that:

- First: it is an element as each sphere seems to be the same.

- Second: It is a compound as two different type of spheres are shown, white and black, considering each sphere is a different element.

- Third: it is an element because the spheres are equal, in fact, it is a diatomic element as two atoms are joined per molecule.

- Fourth: It is a compound as well since each molecule has two colored spheres and one white sphere, meaning that it has two different atoms per molecule.

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4 years ago
What was the biggest challenge you faced in getting to where you are today and how did you overcome it?
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4 0
3 years ago
At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

          \Delta G^{o} = \Delta H^{o} - T \Delta S^{o}    

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
1 When a chemical reaction occurs A. the substances involved do not mix together and maintain their original properties. B. the
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Answer:

D

Explanation:

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6 0
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in the structure of an atom why are protons present in the centre and are not pulled outside by the electrons as both are positi
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Answer:

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