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viktelen [127]
3 years ago
10

PLZ HELPPPP TIMED TESTT!!!!!!?????

Chemistry
1 answer:
Leya [2.2K]3 years ago
5 0

Answer:

my probably best answer is B

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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
2 years ago
Read 2 more answers
The chemistry teacher asked George to transfer an ideal gas from container A to container B. Container B is half the size of con
Rzqust [24]
I think the answer is increase; increase

7 0
3 years ago
Read 2 more answers
An ion has 14 neutrons, 10 electrons, and a mass number of 27 what element
nasty-shy [4]

The ion is Al³⁺

mass number - number of neutrons= atomic number

27 - 14 = 13

Aluminum has an atomic number of 13, thus we know this is the metal in question. Also, because the aluminum has only 10 electrons, (3 less than a neutral atom of aluminum would have), its charge must be 3+

4 0
3 years ago
If ine mole of pennies were divided amung 250 million peoplein
Airida [17]

Answer:

2.4088\times 10^{13} dollars each person will receive.

Explanation:

Number of people in which 1 mole of pennies is distributed = 250 million =

1 million = 10^6

250 million = 2.5\times 10^8 persons

Number of pennies in 1 mole = 6.022\times 10^{23}

Pennies per person:

\frac{6.022\times 10^{23} pennies}{2.5\times 10^8 persons}=2.4088\times 10^{15} pennies/person

1 penny = 0.01 $

2.4088\times 10^{15} pennies/person=2.4088\times 10^{15}\times 0.01 \$/person=2.4088\times 10^{13} \$/person

2.4088\times 10^{13} dollars each person will receive.

7 0
2 years ago
what are we/what has someone did to prevent/lessen the negative side effects of aluminium extraction?
kogti [31]

Aluminum is one of the main factors that reduce plant growth in acid soils. Although it is generally harmful to plants in soils with a neutral medium, the concentration of positive aluminum ions in acid soils increases and malfunctions in root and function growth.

Most acid soils are saturated with aluminum rather than hydrogen ions. Soil acidity is the result of hydrolysis of aluminum compounds. This principle (lime correction) to determine the degree of base saturation in the soil has become the basis of the methods used in soil testing laboratories to determine the lime requirements for soil. Application of lime to soil reduces the toxicity of aluminum to plants. Note This connector loads slowly.

Adaptation of wheat to allow aluminum to be carried out is due to the fact that aluminum releases organic compounds that in turn combine with harmful aluminum cations. It is believed that sorghum has the same endurance. The first genes found to withstand aluminum were found in wheat. Aluminum sulphide bearing has been found to be governed by an individual gene, such as in wheat. This is not the case in all plants.

7 0
3 years ago
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