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oksano4ka [1.4K]
3 years ago
11

What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00

MM ? Express your answer to two significant figures and include the appropriate units.

Physics
1 answer:
pentagon [3]3 years ago
8 0

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

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A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.
GrogVix [38]

Answer:

H = 532 m

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

d = vt

d = 340 \times 3

d = 1020 m

now in the same time the distance that teacher will fall down is given as

d_1 = \frac{1}{2}gt^2

d_1 = \frac{1}{2}(9.81)(3^2)

d_1 = 44.1 m

now total distance traveled by teacher and sound in 3 s

d_{total} = d + d_1

d_{total} = 1020 + 44.1

this total distance must be equal to twice the height of the cliff

2H = 1064.1 m

H = 532 m

7 0
3 years ago
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What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

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Answer:

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Explanation:

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Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

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Explanation:

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Answer:

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Explanation:

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