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3 years ago

7

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

1 answer:

mafiozo [28]3 years ago

5 0

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

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How is a magnetable to pick up bits of metal without actually touching them?

lisov135 [29]

C.There is an invisible magnetic field around the magnet where it exerts magnetic force

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3 years ago

Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A

Finger [1]

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

$r=(100\cos2\theta)\ m$

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

$v=r\omega$

$\omega=\dfrac{v}{r}$

Where, v = velocity

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Put the value into the formula

$\omega=\dfrac{29.0576}{100\cos30}$

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Hence, The rate at which radar must rotate is 0.335 rad/s.

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3 years ago

A student walks 10 m North, 6 m South, 7 m West and 4 m East. Find the student's distance AND displacement

BlackZzzverrR [31]

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Hope this helped!

Mark Brainliest if you feel like it!

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3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.51 m and can rotate about the vertical axis through its center. A 67.

blondinia [14]

Answer:

I = 274.75 kg-m²

Explanation:

given,

mass of horizontal platform = 129 Kg

radius of the disk = 1.51 m

mass of the person(m_p) = 67.5 Kg

standing at distance(r_p) = 1.09 m

mass of dog(m_d) = 25.3 Kg

sitting near the person(r_d) = 1.37 m from center

moment of inertia = ?

Moment of inertia of disc = $\dfrac{1}{2}MR^2$

Moment of inertia of the system

$I = MOI\ of\ disk + m_p r_p^2 + m_d r_d^2$

$I = \dfrac{1}{2}MR^2 + m_p r_p^2 + m_d r_d^2$

$I = \dfrac{1}{2}\times 129 \times 1.51^2 + 67.5 \times 1.09^2 + 25.3\times 1.37^2$

I = 274.75 kg-m²

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3 years ago

Hello please help i’ll give brainliest

balu736 [363]

D, I hope this is correct!

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3 years ago