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masha68 [24]
2 years ago
5

The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l

ong will it take the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm?
Physics
1 answer:
Feliz [49]2 years ago
7 0

Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

After, four half-lives, pressure reduces to 0.00625 atm

Time taken = 4 * 200 = 800 seconds

Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

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When a 4.32 kg object is hung vertically on a certain light spring that obeys Hooke's Law, the spring stretches 2.92 cm.
Alika [10]

(a) 1.01 cm

First of all, we need to find the spring constant of the spring.

The force initially applied to the spring is equal to the weight of the block hanging on it:

F=mg=(4.32)(9.8) = 42.3 N

where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.

When this force is applied, the spring stretches by

x=2.92 cm = 0.0292 m

We can find the spring constant by using Hooke's law:

F=kx

where k is the spring constant. Solving for k,

k=\frac{F}{x}=\frac{42.3}{0.0292}=1448.6 N/m

Later, the first object is removed and another object of mass

m' = 1.50 kg

is hung on the spring. The weight of this object is

F'=m'g=(1.50)(9.8)=14.7 N

So, if we use Hooke's law again, we can find the new stretching of the spring:

x'=\frac{F'}{k}=\frac{14.7}{1448.6}=0.0101 m = 1.01 cm

(b) 1.16 J

The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:

W=\frac{1}{2}kx^2

where we have

k = 1448.6 N/m is the spring constant

x = 4.00 cm = 0.04 m is the new stretching

Solving the equation, we find the work that must be done by the external force:

W=\frac{1}{2}(1448.6)(0.04)^2=1.16 J

6 0
3 years ago
James has a mass of 98 kg and Basma has a mass of 59 kg. James is running at 3.0 m/s, while Basma is running at 4.0 m/s.
masha68 [24]

(a) James has the most momentum which is 294 kgm/s.

(b) The resultant force acting on Basma is 90.78 N.

(c) The time taken for James to stop is 3.2 seconds.

<h3>Momentum of each person</h3>

Momentum of James: P = mv = 98 x 3 = 294 kgm/s

Momentum of Basma: P = mv = 59 x 4 = 236 kgm/s

<h3>Resultant force of Basma</h3>

F = ma = mv/t = P/t = 236/2.6 = 90.78 N

<h3>Time for James to stop</h3>

F = P/t

t = P/F

t = 294/90.78

t = 3.2 s

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

4 0
2 years ago
Match the term with its description.
Mrac [35]

A=Ion

B=pH

C=Acid

D=Alkaline

6 0
3 years ago
Can you hellp me please
mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0
3 years ago
Read 2 more answers
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
  • Resistance=R=210Ohm
  • Power=9.28W=P

Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}

\\ \sf\longmapsto I\approx\dfrac{2}{10}

\\ \sf\longmapsto I\approx0.2A

4 0
3 years ago
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