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3 years ago

A hydrogen ion, H+, is the same as a.... PLEASEEE

1 answer:

5 0

It is the same as a proton. That is why, if hydrogen was a molecule not an ion, one electron is needed to 'cancel' out the proton.

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

Ill give brainliest :D<br><br> how many grams of kcl are produced from 11.5 grams?

eduard

Answer:

approximately 15.1 grams.

Explanation:

The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.

change 25 grams of potassium chlorate to moles.

calculate the gram molecular mass of potassium chlorate.

Chlorate is Cl with 3 oxygens. ate = saturated. Chlorine has seven valance electrons when it is saturated six of these electrons are used by oxygen ( 2 electrons per oxygen) leaving only 1 electron.

1 K x 39 grams/mole

+1 Cl x 35.4 grams/ mole

+3 O x 16 grams/ mole

= 122.4 grams / mole Potassium Chlorate

122.4

= moles.

2.05 moles of Potassium Chlorate.

There is a 1:1 mole ratio. 1 mole of Potassium Chlorate will produce 1 mole of Potassium Chloride.

2.05 moles of Potassium Chlorate will produce 2.05 moles of Potassium Chloride.

Find the gram molecular mass of Potassium Chloride.

1 K x 39 = 39

+1 Cl x 35.4 = 35.4

= 74.4 grams / mole.

2.05 moles x 74.4 grams/ mole = 15.2 grams

4 0

3 years ago

Read 2 more answers

Which of the following trends is indirectly proportional to effective nuclear charge, Zeff

Julli [10]

Answer : Option C) Atomic Size

Explanation : The atomic radius of the elements is found to be decreasing if we go from left to right in the modern periodic table. Accordingly, $Z_{eff}$ increases as the number of shielding electrons present in the atomic nucleus of the periodic elements which lies in the same row remains constant while the number of protons in each atomic shell increases.

The effective nuclear charge $Z_{eff}$ of an atom is defined as the net positive charge which is felt by the valence electron of the atomic element.

When $Z_{eff}$ is observed to decrease, it is seen that the atomic radius grows in size. So, it explains the inverse relationship between both. This phenomenon occurs, because there is more screening of the electrons from the nucleus taking place, which is observed due to decrease the attraction between the electron and the nucleus.