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Kisachek [45]
2 years ago
5

Liquid water boiling in an open pan on a stove is our system. Is the system open or closed? What boundary transfers are happenin

g? Is the system at steady state?
Chemistry
1 answer:
valina [46]2 years ago
3 0

Answer:

It's an open system, tranfering heat through a rigid, diathermal wall and matter through an imaginary and permeable wall, and it is not at steady state.

Explanation:

  • An <em>open system</em> is that that interacts with its surroundings exchanging energy and matter. In an open pan with boiling water you have an open system because steam (matter) is leaving the system, as well as heat (energy) through the pan/stove.
  • A<em> boundary</em> is what separates the system from its surroundings, there are many types of boundaries, based on how they transfer energy they can be diathermal (conducting heat) or adiabatic (insulating), on their rigidity they can be rigid, flexible, imaginary or movable and based on their permeability. For the system described we have an imaginary boundary on top that is also permeable allowing matter to go out or in the system, and another wall (the stove/pan itself that is rigid and impermeable avoiding the loss of matter and diathermal, allowing the conduction of heat.
  • It is said that a system is at a<em> steady state</em> when the variables that define that system remain constant over time. In an open pan, you can't fully control those variables, you'll have matter and energy scaping from it with no way to regulate it.

I hope you find interesting and useful this information! good luck!

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Aleonysh [2.5K]

The gold leaf (not foil) is to give a sculpture or a painting appeal. Without it, it would discolour quicker.The gold leaf can be easily removed if needed to be.

5 0
3 years ago
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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

3 0
3 years ago
A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete
lys-0071 [83]

ke prop to v^2


ke1/v1^2=ke2/v2^2


400/50x50=joules/100x100


400x2x2


1600j

8 0
3 years ago
If an atom with a mass of 47 amu has 22 neutrons, what atom is it?
LenaWriter [7]

Answer:

Manganese

Explanation:

At Mass - No neutrons = Atomic Number = #protons in nucleus

47 - 22 = 25 => At. No. 25 is Manganese (Mn)

7 0
2 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

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

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
3 years ago
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