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3 years ago

6

Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive δg at rt be made f

avorable by increasing the reaction temperature?

1 answer:

Over [174]3 years ago

7 0

I believe that the answer to the question provided above is yes, an unfavorable reaction that has a positive δg at rt be made favorable by increasing the reaction temperature.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

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Answer:

b

Explanation:

bbbbbbbbbbbbbbvgh c tyvftj xf

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2 years ago

Knowledge and skills learned through socialization are an example of

ziro4ka [17]

Answer:

I think no.2 the answer

Because socialization and social resources are both for me

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3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

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3 years ago

Hlo <br><br> what is a force........??

Paha777 [63]

Explanation:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.

Formula

Newton's Second Law

F = m * a

F = force

m = mass of an object

a = acceleration

5 0

2 years ago