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polet [3.4K]
2 years ago
7

Calculate the energy difference for a transition in the paschen series for a transition from the higher energy shell n=4. expres

s your answer to four significant figures and include the appropriate units.
Chemistry
1 answer:
Troyanec [42]2 years ago
5 0
Electrons are orbiting around the nucleus in a specific energy level as described in Bohr's atomic model. There are 7 energy levels all in all; 1 being the strongest and nearest to the nucleus, and 7 being the weakest and farthest away from the nucleus. Electron can transfer from one energy level to another. If it increases energy, it absorbs energy. If it goes down an energy level, it emits energy in the form of light. This light can be measure in wavelength through the Rydberg equation:

1/λ =R(1/n₁² -1/n₂²), where
λ is the wavelength
R is the Rydberg constant equal to 1.097 × 10⁻7<span> per meter
n</span>₁ and n₂ are the energy levels such that n₂>n₁

In the Paschen series is an emission spectrum of hydrogen when the energy level is at least n=4. So, this covers n=4 to n=7.

1/λ =(1.097 × 10⁻7)(1/4² -1/7²)
λ  = 216.57 ×10⁻⁶ m or 216.57 μm
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2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:
Nesterboy [21]

Answer

  • increase in temperature
  • decrease in pressure
  • continuous removal of PH3
  • adding more of P into the system

Explanation:

        In the reaction   P4(g)+6H2(g) ⇌ 4PH3(g);

  • The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

        heat+A⇌BΔH=+

  • Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
  • According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0
2 years ago
Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.
andrezito [222]

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

5 0
2 years ago
Which gas tank will empty first? acetylene (C2H2) oxygen (O2)
Monica [59]
The answer is oxygen. (02)
4 0
2 years ago
Read 2 more answers
What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?
Ilya [14]

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

4 0
2 years ago
7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25
horsena [70]

Answer:

The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.

Explanation:

Mass of compound X = 7.00 g

Moles of compound X = \frac{7.00 g}{70 g/mol}=0.100 mol

Mass of water in calorimeter ,m= 35.00 kg = 35000 g

Change in temperature of the water in calorimeter = ΔT

ΔT = 2.113°C

Specific heat capacity of water ,c= 4.186 J/g °C

Q =  m × c × ΔT

Q=35000 g\times 4.186 J/g ^oC\times 2.113^oC=309,575.6 J=309.575 kJ

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.

Heat of formation of Compound X at 25°C:

\frac{-Q}{\text{moles of compound X}}=\frac{-309.575 }{0.100 mol}

= -3095.75 kJ/mol

6 0
2 years ago
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