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:
a2 = 3a1
Explanation:
Detailed explanation and calculation is shown in the image below
Input= ?
Output= 0.75
MA= 0.33
So I just kept on Dividing numbers until I was close to "0.33" & I cam up with the answer of .......... 0.25.
So when you divide 0.25 by 0.75 you get the MA of 0.33
Answer:
Neutrally charged!!!!!!!!!!!!!!!!!!!!!
Explanation:
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
