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DIA [1.3K]
2 years ago
14

Which statement is true for an open system?

Chemistry
1 answer:
Bingel [31]2 years ago
3 0

Answer:

Before we get into the first law of thermodynamics we need to understand the relation between heat and work and the concept of internal energy. Just like mass, energy is always conserved i.e. it can neither be created nor destroyed but it can be transformed from one form to another. Internal energy is a thermodynamic property of the system that refers to the energy associated with the molecules of the system which includes kinetic energy and potential energy.

Whenever a system goes through any change due to interaction of heat, work and internal energy, it is followed by numerous energy transfer and conversions. However, during these transfers, there is no net change in the total energy.

Similarly, if we look at the first law of thermodynamics it affirms that heat is a form of energy. What it means is that the thermodynamic processes are governed by the principle of conservation of energy. The first law of thermodynamics is also sometimes referred to as the Law of Conservation of Energy

Explanation:

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CuSO^4 is an example of a(n)_____.
GarryVolchara [31]

Answer:

Compound.

Explanation:

That would be compound as it consists of a number of elements bonded together.  It is inorganic not organic.

3 0
3 years ago
The specific heat of wood is 2.03 J/g.°C. How much
Brilliant_brown [7]

Answer:

11419 J/g/ 11.419 KJ/g

Explanation:

H=MCQ

H=225×2.03×(-15-10)

H=225×2.03(25) Note; negative sign is of no use

H=11419J/g

3 0
3 years ago
Given what you observe at the molecular/atomic level, why would surface area have this effect?
LenKa [72]

Answer:

Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate. Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur.

Explanation:

:)

5 0
2 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
Why do you think plant cells need a cell wall as well as a cell membrane?
Sergio039 [100]

Answer:

to protect it from harm

Explanation:

3 0
3 years ago
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