Answer:
The pressure of the gas at 23 C is 179.92 kPa.
Explanation:
Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
In short, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.
Gay-Lussac's law can be expressed mathematically as follows:
Studying two states, one initial 1 and the other final 2, it is satisfied:
In this case:
- P1= 310 kPa
- T1= 237 C= 510 K (being 0 C= 273 K)
- P2= ?
- T2= 23 C= 296 K
Replacing:
Solving:
P2= 179.92 kPa
<u><em>The pressure of the gas at 23 C is 179.92 kPa.</em></u>
Answer:
The answer to your question is below
Explanation:
Material Classification Reason
Crude oil Mixture Because we know that crude oil
is composed of different substances.
Butane Compound Because it is composed of carbon
and hydrogen
Dodecane Compound Because it is composed of carbon
and hydrogen.
Octane Compound Because it is composed of carbon and
hydrogen.
Benzene Compound It is also composed of carbon and
hydrogen.
Gasoline Mixture It is a combination of hydrocarbons
Kerosene Mixture It is a combination of hydrocarbons
Oxygen Element We can find it in the periodic table.
Carbon dioxide Compound It is composed of carbon and oxygen.
Water Compound It is composed of hydrogen and
oxygen.
Answer:
893 moles
Explanation:
An ideal gas at STP occupies 22.4 liters. Calculating Oxygen as if it were an ideal gas there are . 893 moles of Oxygen in 20.0 liters.
Answer:
3.1°C
Explanation:
Using freezing point depression expression:
ΔT = Kf×m×i
<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>
Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:
9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m
Replacing in freezing point depression formula:
ΔT = 5.12°cm⁻¹×0.472m×1
ΔT = 2.4°C
As freezing point of benzene is 5.5°C, the new freezing point of the solution is:
5.5°C - 2.4°C =
<h3>3.1°C</h3>
<em />
Bonding between 2 hydrogen atoms is 'covalent bond'.
Hope this helps you.
