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2 years ago

7

The rate constant for this second‑order reaction is 0.760 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for

the concentration of A to decrease from 0.750 M to 0.330 M?

1 answer:

Mashcka [7]2 years ago

3 0

Answer:

2.23 s

Explanation:

For a second-order reaction:

1 / [A] = 1 / [A]₀ + kt

Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:

1 / 0.330 = 1 / 0.750 + 0.760t

t = 2.23

It would take 2.23 seconds.

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There is a limit to how long your neck can be. If your neck were too long, no blood would reach your brain! What is the maximum

spayn [35]

Answer:

The maximum height a person's brain could be above his heart is: 1.28 meter.

Explanation:

We need to know what is the normal blood pressure ours hearts so there is a rate: 120/80 (mmHg) and the average will be: 100 (mmHg) and using the Pascal law that relate pressure, density, gravity and height like: $P_{2} = pgh_{1} - pgh_{2} + P_{1}$ , where P is pressure, p is density, g is the gravity acceleration and h is the height. Now we can find the height and delta of pressure will be: P2-P1 = 100 (mmHg), knowing that 1(mmHg) is equal to 133 Pa, we can do the convertion to 13332.2 (Pa), now because the units of Pascal are Newton/(meter^2). Then we solve the formula to get the height: $\frac{P2-P1}{pg} =h$ so we get: $\frac{13332.2}{(1060*9,81)}=Height=1.28(meters).$

5 0

3 years ago

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0

3 years ago

Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago

D. What is the net force on the bowling ball rolling lane

3241004551 [841]

Answer:

Friction

Explanation:

3 0

2 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago