<span>A fast moving stream of air has a lower air pressure than a
slower air stream. As the stream of air moved over the
top of the paper, the air pressure over the paper dropped. The
air pressure underneath the paper stayed the same. The
greater air pressure underneath lifted the paper strip and it
rose. The idea that a moving air stream has lower air pressure
than air that is not moving is called “Bernoulli’s Principle”.
</span>The
force of the moving air underneath the balloon was enough to
hold it up. The weight added by the paper clip prevents
the balloon from going too high. But that is only part
of the story. The balloon stays inside the moving stream
of air because the pressure inside is the air stream is lower
than the still air around it. As the balloon moves toward the
still air outside of the air stream, the higher pressure of
the still air forces the balloon back into the lower pressure
of the air stream. Bernoulli’s Principle at work again!
Answer:
Reaction 1: Kc increases
Reaction 2: Kc decreases
Reaction 3: The is no change
Explanation:
Let us consider the following reactions:
Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol
Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol
Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol
To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>
Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.
Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.
Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.
The number of moles of oxygen required to generate 28 moles of water from the reaction is 14 moles
<h3>Balanced equation </h3>
2H₂ + O₂ —> 2H₂O
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
<h3>How to determine the mole of oxygen needed </h3>
From the balanced equation above,
2 moles of water were obtained from 1 mole of oxygen
Therefore,
28 moles of water will be obtained from = 28 / 2 = 14 moles of oxygen
Thus, 14 moles of oxygen are needed for the reaction
Learn more about stoichiometry:
brainly.com/question/14735801
In chemistry:
When something is oxidized, it loses electrons.
When something is reduced, it gains electrons.
So the correct statements would be the second to last and third to last statements
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.
