Complete Question
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices
Answer:
Explanation:
From the question we are told that:
Wavelength 
Generally the equation for Atom Transition is mathematically given by
Where
Rydberg constant 
Therefore
<h2>Answer:</h2>
At 189 K volume = 32.0 cm
As volume of gas is directly proportional to temperature.
As the temperature increase the volume of gas will also increase.
so
Volume at 1 K = 32/189
Volume at 242 K = 32/189 * 242 = 40.97 cm.
Gas occupy 40.97 cm volume at 242 K.
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
Answer:
Cu(s) + N₂(g) + 2O₂(g) ⟶ Cu(NO₂)₂(s)
(a) 43.6 mg; (b) 520 mg
(a) <em>Mass of phosphoric acid (PA) in a dose
</em>
Mass of PA = 2 tsp × (21.8 mg PA/1 tsp) = 43.6 mg PA
(b) <em>Mass of PA in the bottle
</em>
<em>Step 1</em>. Convert <em>ounces to millilitres
</em>
Volume = 4 oz × (30 mL/1 oz) = 120 mL
<em>Step 2.</em> Calculate the mass of PA
Mass of PA = 120 mL × (21.8 mg PA/5 mL) ≈ 520 mg PA
