A net force is applied on a 100 kg rocket which causes the rocket to acceleration at 10 m/s2. The same net force is applied on a
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Answer:
The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;
(b) 4, 3, -3, 1/2.
Explanation:
To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;
4, 4, 4, 1/2
4, 3, -3, 1/2
4, 3, 0, 0
4, 5, 7, -1/2
4, 4, -5, 1/2
Let us label them as a to as follows
(a) 4, 4, 4, 1/2
(b) 4, 3, -3, 1/2
(c) 4, 3, 0, 0
(d) 4, 5, 7, -1/2
(e) 4, 4, -5, 1/2
Next we note the rules for the assignment and arrangement of quantum numbers are as follows
Number Symbol Possible values
Principal Quantum Number .......n........................1, 2, 3, ......n
Angular momentum quantum
number...............................................l.........................0, 1, 2, .......(n - 1)
Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l
Spin Quantum Number.................m.....................+1/2, -1/2
We are meant to analyze each of the arrangement for acceptability.
Therefore for (a),
we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.
Therefore (a) is not acceptable.
(b) Here we note that
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.
(c) Here we have
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable
The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 0 ∉ (+1/2, -1/2) → not acceptable
Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.
(d) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = -1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
(e) Here we have;
The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable
The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable
The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable
The spin quantum number m = 1/2 ∈ (+1/2, -1/2) → acceptable
Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.
Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
- Heat loss = Heat gained
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = = 3348.8J/kg C≅ 3349J/kg C