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vodomira [7]
3 years ago
8

Select the correct answer.

Physics
1 answer:
chubhunter [2.5K]3 years ago
3 0
I believe it’s self-referent encoding
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The compound PCl5 decomposes into Cl2 and PCl3. The equilibrium of PCl5(g) Cl2(g) + PCl3(g) has a Keq of 2.24 x 10-2 at 327°C. W
nordsb [41]
Take note of the reaction formula which is PCl5=Cl2+PCl3.
The Keq = [Cl2] * [PCl3] / [PCl5]=2.24*10^-2.
For the reason that the volume is 1 liter, the concentration of Cl2 will be computed through: <span>(2.24 * 10^-2) * 0.235 / 0.174 </span> = 0.0303 mol/L is the answer.
7 0
3 years ago
Read 2 more answers
How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.
liraira [26]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

4 0
3 years ago
Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

8 0
3 years ago
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