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3 years ago

11

For the hypothetical reaction A → B, calculate the average rate of disappearance of A if the initial concentration of A is 0.91

M and the concentration of A after 90 minutes is 0.11 M.

1 answer:

marishachu [46]3 years ago

7 0

Answer:

The right answer is 8.9 x 10^-3 M/min

Explanation:

A → B

-d [A]/dt = K [A]

ΔA/Δt = - (C2 -C1)/t2 - t1

= - (0.11 - 0.91)/90

= 8.9 x 10^-3 M/min

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What is the speed of a truck traveling 10km in 10 minutes

user100 [1]

Answer: 58.8235 km/h

speed = distance/time

the distance is 10 km

the time is 10 minutes

the unit is not correct, so we first change minute to hour

so 10/60 is 0.166667, rounded to 0.17.

10 km/ 0.17 hours =

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1 year ago

The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving

nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

$F_f = \mu_s (mg)$

where

$\mu_s = 0.5645$ is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

$F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N$

So, the applied force must be greater than this value.

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3 years ago

"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3

avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by $B=\frac{\mu _0i}{2\pi r}$

It is given that magnetic field is zero at some distance so

$\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}$

Here $i_1=3\ A \ and\ i_2=8\ A$

So $\frac{3}{x}=\frac{8}{24-x}=6.5454\ m$

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3 years ago

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</span>

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3 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

3 years ago