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3 years ago

Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

4 0

Explanation:

Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol Q.

The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol K.

The relationship between Gibbs free energy change and reaction quotient of the reaction is:

$\Delta G=\Delta G^o+RT ln Q$ ......(1)

where,

$\Delta G$ = Gibbs free energy change

$\Delta G^o$ = Standard Gibbs free energy change

R = Gas constant

T = Temperature

At equilibrium, the free energy change of the reaction becomes 0 and standard Gibbs free energy change can be related to the equilibrium constant by the equation:

$\Delta G^o=-RT ln Q$ ...(2)

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S_A_V [24]

The speed of the stone is 7.7 m/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically,

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the stone in this problem, we have:

K = 150 J is its kinetic energy

m = 5.0 kg is its mass

Re-arranging the equation for v, we find the speed of the stone:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(150)}{5.0}}=7.7 m/s$

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3 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

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3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

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