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Mamont248 [21]
2 years ago
5

A hydrogen atom is in its fifth excited state. The atom emits a wavelength photon. Determine the maximum possible orbital angula

r momentum of the electron after emission. Express your answer as multiples of hbar
Chemistry
1 answer:
forsale [732]2 years ago
8 0

Answer:

(√ 20) hbar.

Explanation:

The first thing to do is to calculate the second principal quantum number, n from the formula for Calculating energy,E and this is done as given below;

Energy, E = 13.6 (1 / n^2 - 1/6).

(Where energy,E = 1240/ 7474).

Hence, when we solve the above equation for n, the principal quantum number = 5.

We know that the azimuthal quantum Number, l = (n- 1); 5 - 1 = 4.

Therefore, the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 4 × 5) hbar.

the maximum possible magnitude of the orbital angular momentum of the atom after emission = (√ 20) hbar.

Note that hbar = 1.0545718 × 10-34 m2 kg / s.

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A student needs exactly 1 gram of baking powder to conduct an experiment. The balance reads 0.37 grams. How
sammy [17]

Answer:

B

Explanation:

If the student needs one gram but so far only has 0.37 grams, then the amount they need is the difference between what they need and how much they already have. 1-0.37=0.63 grams.

...which isn't actually an option because none of them have decimal points but I would say it is B anyway because it is the equivalent ratio and maybe there was a typo.

Hope this helped!

6 0
3 years ago
Read 2 more answers
What is the vapor pressure of an aqueous solution made up of 60.0g of urea (CH4N2O) in 180g of water? the vapor pressure of wate
djyliett [7]
To get the answer you use the Law of Raoult.


Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.


ΔP = Pa * Xa


Here Pa = 0.038 atm


And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b


Na = mass of urea / molar mass of urea =  60 g / (molar mass of CH4N2O)


molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol


Na = 60 g / 60 g/mol = 1 mol


Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol


Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091


ΔP = Pb * Xa = 0.038 atm * 0.09091 =  0.0035 atm


Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.


 Answer: 0.035 atm
7 0
3 years ago
A. Clearly draw the Lewis structure for the PBr4- ion. Show your math where
Nataliya [291]

Answer:

   Br

    |

Br-P-Br

    |

   Br

Explanation:

To calculate the valance electrons, look at the periodic table to find the valance electrons for each atom and add them together. P is in column 5A, so it has 5, Br is in column 7A, so it has 7 (multiply by 4 since there are 4 Br atoms to give 28) and there is a 1- charge, so add one more electron. 5+28+1=34, so there are 34 electrons to place. P would be the central atom, so place it in the middle. Place each Br around the P (as shown above) with a a single line connecting it. Each line represents 2 electrons, so 8 total have been place, leaving 26 remaining. Place 6 electrons around each Br (2 on each of the unbonded sides), which leaves 2 electrons remaining. The remaining pair of unbound electrons will be attached to the P between any two Br atoms. Phosphorus doesn't have to follow the octet rule, so it actually ends up with 10 valance electrons.

4 0
2 years ago
Read 2 more answers
You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol
sashaice [31]

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

8 0
3 years ago
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
3 years ago
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