If
SO3(g)
is removed from the following reaction, will the equilibrium shift to the left, shift to the right, or stay the same? Explain.
2SO2(g)+O2(g)⇋2SO3(g);ΔH
Explanation: The reaction shown in the question is a combination reaction between sulfur dioxide gas and oxygen gas, forming sulfur trioxide gas by the two gases combining into one product. The question's objective is to determine the direction in which the equilibrium will shift if sulfur trioxide is removed. Removing the products from the container during a reversible chemical reaction means that only the forward reaction will proceed right after the products are removed. Once more of the products are formed, the reverse reaction will start to occur.
But, when the product is removed, the system will compensate for the removal of the product by increasing the production of the product, which is done by increasing the rate of the forward reaction and shifting the equilibrium to the right.
Answer:
When it is marked with its cubic-inch volume
Explanation:
Because this allows for best and efficient identification
Answer:
Explanation:
Remark
In general, these 3rd class levers are very inefficient. Because the force distance is smaller than the load distance, you need to pull upward with more force that the weight of the load. So whatever the load is, the force is going to be much greater.
The distances are always measured to the pivot unless you are asked something specific otherwise.
Givens
F = ?
weight = 6N
Force Distance = F*d = 0.5 m
Weight Distance =W*d1 = 2 m
Formula
F*Fd = W*Wd
Solution
F*0.5 = 6 * 2 Divide by 0.5
F = 12/0.5
F = 24 N upwards
Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time
Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>
Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration
Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N
<em>ANSWER: 240 N </em>
Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²
Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m
<em>ANSWER: d = 13.5 m</em>
