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yanalaym [24]
3 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?
Physics
1 answer:
Mashcka [7]3 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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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0
4 years ago
How to calculate energy needed for a change of state ​
Alenkasestr [34]

Answer:

change in temperature = (100 - 25) = 75.0°C.

change in thermal energy = mass × specific heat capacity × change in temperature.

= 0.200 × 4,180 × 75.0.

= 62,700 J (62.7kJ)

7 0
3 years ago
Read 2 more answers
A box rests on a horizontal, frictionless surface. a girl pushes on the box with a force of 17 n to the right and a boy pushes o
Komok [63]
Refer to the diagram shown below.

The net force acting on the box is 17 - 13 = 4 N to the right.
The box moves on a friction surface by 3.5 m to the right.

By definition,
Work = Force x Distance.

(a) The work done by the girl is
     W₁ = (17 N)*(3.5 m) = 59.5 J

(b) The work done by the boy is
     W₂ = (13 N)*(-3.5 m) = - 45.5 J

(c) The work done by the net force  is
    W₃ = (4 N)*(3.5 m) = 14 J

Note that W₃ = W₁ + W₂

Answers:
(a) 59.5 J
(b) - 45.5 J
(c) 14 J

5 0
4 years ago
The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha
Marizza181 [45]

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

8 0
4 years ago
All stars go through a lifecycle.
Komok [63]

Mira is much bigger than the Sun.

Only very massive stars will go through a supernova stage, causing the outer layer to explode away and the core to collapse in on itself, becoming very dense.

5 0
2 years ago
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