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PtichkaEL [24]
3 years ago
10

What is the equilibrium concentration of ibr?

Physics
1 answer:
asambeis [7]3 years ago
3 0
IBR is the thermal decomposition of iodine(I) bromide to produce iodine and bromine. This reaction takes place at a temperature of over 40,5°C and is written as: 

 <span>2IBr ⇄ I2 + Br2
</span>
Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.<span> You can calculate the equilibrium concentration if you know the equilibrium constant Kc (Kc=I^2*Br^2/IBR^2) and the initial concentration for the reaction. The initial concentration is obtained from ICE Table.</span>
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Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
Free_Kalibri [48]

Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

So if n= 5 it can take 0,1,2,3,4

i.e it can take 5 values

2)for an electron with l =3

it can be from -3 -2 -1 0 1 2 3

i.e it can take 7 values

3) n = 3 !!

l = 0 , 1 , 2

for l=0 , m = 0 total = 1

for l= 1 ,m = -1,0,1 total = 3

for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

total possible states = 9 * 2 = 18

Answer is 168

4)given l=3 and n=3

orbital quantum number cannot be equal to principal quantum number

its max value is l-1 only

5)L = sqrt(l(l+1))x h'

for it to be max l should be max

for n = 3 max l value is 2

therfore it is sqrt(2(2+1)) x h'

\sqrt(6) \times h'

this is the answer

5 0
3 years ago
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
A cannon elevated at 40 degrees is fired at a wall 300m away on level ground. The initial speed of the cannonball is 89m/s How l
lianna [129]
For the answer to the question above, we'll have to use these formulas.

A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40 

<span>B) y=vtsin 40 - gt^2/2, just sub in
</span>
I believe you can do the rest.

I hope I helped you with my answers.
3 0
3 years ago
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
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