Ask question

Ask question

All categories

3 years ago

15

All of the statements are true for spontaneous reactions except: a) the reaction rate is determined by the value of ΔG. b) if th

e enthalpy change is unfavorable, they occur at a high temperature. c) the value of ΔG is unaffected by a catalyst. d) the value of ΔG is less than zero. e)they are said to be exergonic.

1 answer:

Dahasolnce [82]3 years ago

5 0

Answer:

A is false

Explanation:

A is false.

● activation energy affects the rate of a reaction and catalyst affects the rate of reaction too

ΔG (change in free energy) tells if a reaction is spontaneous or not

You might be interested in

If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m, what minimum f

Marina86 [1]

We have that the Force is mathematically given as

F=170.833N

<h3></h3><h3>Force </h3>

Question Parameters:

The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,
the end of a 24 cm-long wrench to <em>loosen </em>the nut

Generally the equation for the Force is mathematically given as

$F=\frac{T}{d}$

Therefore

F=41/0.24

F=170.833N

For more information on Force visit

brainly.com/question/26115859

3 0

2 years ago

Need help please... thank u

tangare [24]

Answer:

the answer from my side is both vertical and horizontal

4 0

3 years ago

Read 2 more answers

What is the average velocity of the person taking this walk?

Butoxors [25]

5 would be the answer

6 0

3 years ago

Read 2 more answers

Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride.

Mashcka [7]

The amount of work which Sam does need to stop the boat is 784Joule if boat mass is 800kg and boat drift in 1.4m/sec.

We know that according to work-energy theorem, change in kinetic energy of the body from one speed to another is equivalent to amount of work done by all forces acting on the body.

So, here we can see that final speed of boat is =0m/sec since Sam need to stop it.

Initial speed of boat is 1.4m/sec.

Also, we know that kinetic energy is given by the below formula:

Kinetic energy=(1/2)mv²

where m is the mass of the body

and v is the speed of the body

Now, from work energy theorem, we get

=>W=(1/2)m $v^2_2$ - (1/2)m $v^2_1$

=>W=0-(1/2)×800×(1.4)²

=>W= - 400× 1.4 × 1.4

=>W = -400×1.96

=>W= -784Joule

Here negative sign denotes work has to done against the original motion of boat.

Hence, amount of work needs to be done by Sam is 784Joule.

To know more about work,visit here:

brainly.com/question/18094932

#SPJ4

3 0

1 year ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago