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Kisachek [45]
3 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]3 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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3 years ago
How does radioactive isotopes decay?
aksik [14]
They can decay through one of three ways: 
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7 0
3 years ago
For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l)
juin [17]

Answer:

The maximum amount of CO2 that can be formed is 9.15 grams CO2

O2 is the limiting reactant

There will remain 4.82 grams of benzene

Explanation:

Step 1: Data given

Mass of benzene = 7.53 grams

Mass of oxygen gas = 8.33 grams

Molar mass of benzene = 78.11 g/mol

Molar mass oxygen gas = 32.00 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles C6H6 = 7.53 grams / 78.11 g/mol

Moles C6H6 = 0.0964 moles

Moles O2 = 8.33 grams / 32.00 g/mol

Moles O2 = 0.2603 moles

Step 4: Calculate the limiting reactant

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles

There will remain 0.0964 - 0.0347 = 0.0617 moles benzene

This is 0.0617 moles * 78.11 g/mol = <u>4.82 grams benzene</u>

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Step 5: Calculate moles CO2

For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.208 moles * 44.01 g/mol

<u>Mass CO2 = 9.15 grams</u>

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