Answer:
Explanation:
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.
SO3 (g) + NO (g) U SO2 (g) + NO2 (g)
Initial (M) 2.00 2.00 0 0
Change (M) −x −x +x +x
Equil (M) 2.00 − x 2.00 − x x x
2 2 c 3
2
c 2
[SO ][NO ]
[SO ][NO]
(2.00 )
=
= −
K
x K
x
Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of
NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.
2.00 − x = 1.30
x = 0.70
Substituting back into the equilibrium constant expression:
2c 2 2c 2
(2.00 )
(0.70)
(2.00 0.70)
= − = −
x KxK
Kc = 0.290
Answer:
The correct answer is option 2.
Explanation:
Colligative property is defined as property which depends upon only on the numbers of particles of solute dissolved in definite amount of solvent, It do not depend on the nature of the solute.
For example : NaCl solution with 0.4 molal will show same colligative properties as a that of the glucose solution with 0.04 molal concentration.
The following are the examples of colligative property:
1. Relative lowering of vapor pressure.
2. Osmotic pressure
3. Elevation in boiling points
4. Depression in freezing point
Answer:
Therefore it will take 7.66 hours for 80% of the lead decay.
Explanation:
The differential equation for decay is
Integrating both sides
ln A= kt+c₁
[
]
The initial condition is A(0)= A₀,
.........(1)
Given that the 
has half life of 3.3 hours.
For half life 
putting this in equation (1)
[taking ln both sides, 
]
⇒k= - 0.21
Now A₀= 1 gram, 80%=0.8
and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram
Now putting the value of k,A and A₀in the equation (1)
⇒ t≈7.66
Therefore it will take 7.66 hours for 80% of the lead decay.
