Answer:
For this angular momentum, no quantum number exist
Explanation:
From the question we are told that
The magnitude of the angular momentum is 
The generally formula for Orbital angular momentum is mathematically represented as
Where 
is the quantum number
now
We can look at the given angular momentum in this form as
comparing this equation to the generally equation for Orbital angular momentum
We see that there is no quantum number that would satisfy this equation
<span>In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can neither be created nor destroyed; rather, it transforms from one form to another.</span>
The given sentence is part of a longer question.
I found this question with the same sentence. So, I will help you using this question:
For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before
any reaction occurs?
(a) Q = K<span>; The reaction </span>is at equilibrium.
(b) Q < K<span>;
The reaction </span>will proceed to
the right.
(c) Q > K<span>; The reaction </span>will proceed to the left.
The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left,
since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:
Kp is the equilibrium constant in term of the partial pressures of the gases.
Q is the reaction quotient. It is a measure of the progress of a chemical reaction.
The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.
At equilibrium both Kp and Q are equal. Q = Kp
If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,
If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.
Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.
Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
Answer:
[K₂CrO₄] → 8.1×10⁻⁵ M
Explanation:
First of all, you may know that if you dilute, molarity must decrease.
In the first solution we need to calculate the mmoles:
M = mmol/mL
mL . M = mmol
0.0027 mmol/mL . 3mL = 0.0081 mmoles
These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.
New molarity is:
0.0081 mmoles / 100mL = 8.1×10⁻⁵ M
Answer:
the activation energy Ea = 179.176 kJ/mol
it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
Explanation:
From the given information
Thus; 
Because at 113.0°C; the rate is 7 time higher than at 100°C
Hence:
1.9459 = 
Ea = 179.176 kJ/mol
Thus; the activation energy Ea = 179.176 kJ/mol
b)
here;
where ;
Now;
t = 7.0245 mins
Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
