Describe the difference between a chemical reaction and a physical change.

1 answer:

5 0

A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity.

Examples of chemical changes are burning, cooking, rusting, and rotting.

Examples of physical changes are boiling, melting, freezing, and shredding.

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A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?

hodyreva [135]

Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K
V₂ = (2.00 L / 313 K) x 303 K
V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C is 1.94 L

7 0

3 years ago

Thermal energy always moves from a greater energy level to a lesser energy level

Nookie1986 [14]

Answer:

$\fbox{Law -Based \: on \: laws \: of \: thermodynamic}$

Explanation:

At constant pressure Thermal energy always moves from a greater energy level to a lesser energy level, laws of thermodynamics prove that.

Nature always likes to attain equilibrium either it's movement of heat energy or flow of water from higher region to lower region. The first and second law of thermodynamics are profe of that, the first law says that the total energy of universe is Constant. Energy can not be destroyed it always changes from one form to another, by work and heat. The second law explains why thermal energy moves from a greater energy level to a lesser energy level, it deals with the change in entropy of a system and surrounding and states heat flows from hot environment to cold environment.

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6 0

2 years ago

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to

Rainbow [258]

Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: $\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }$

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

$56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm$