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2 years ago

Sulfur trioxide (SO3) can be produced using the reversible reaction shown.

1 answer:

3 0

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.

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A gyre is a set of currents that form:

tangare [24]

A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>

7 0

3 years ago

Read 2 more answers

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

5 0

2 years ago

Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4

arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

initial velocity of the car, u = 76 m/s

acceleration of the car, a = - 9 m/s²

time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

height above the ground, h = 500 m

velocity of spaceship, u = 150 m/s

time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

$h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) - (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m$

Learn more here: brainly.com/question/24527971

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2 years ago

HELPPPPPPP!!!!!!!!!!!!!!!!

joja [24]

Point b since it gains the most height, gaining the most gravitational potential energy

Hope this helps and good luckk u can do thisss :)

6 0

2 years ago

QuestionDetails:

sleet_krkn [62]

To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

$r = \sqrt{x^2+R^2}$