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vladimir1956 [14]
3 years ago
8

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

46.3 nm is emitted. The energy difference between these 2p and 2s orbitals is:_________a. 3.07 Ã 10^â28 Jb. 3.07 Ã 10^â19 Jc. 3.07 Ã 10^â17 Jd. 1.28 Ã 10^â31 Je. none of these
Chemistry
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

Explanation:

Wavelength of the photon emitted = \lambda =646.3 nm =646.3\times 10^{-9} m

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

E=\frac{hc}{\lambda }

h = Planck's constant = 6.626\tiomes 10^{-34} Js

c = Speed of the light = 3\times 10^8 m/s

E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}

E=3.07\times 10^{-19} J

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

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miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if        1 g is equal to 100 cg

then  3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if         1 km is equal to 1000 m

then   Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

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3 years ago
Riley is doing an experiment working with friction. He wants to find two ways to modify an object that could increase or decreas
Shtirlitz [24]
Riley can either change the surface area of the object or can change the slipperiness of the material.
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What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?
Maru [420]

When sodium amide i.e.NaNH_{2} reacts with water i.e. H_{2}O results in the formation of sodium hydroxide i.e. NaOH and ammonia NH_{3}.

The chemical reaction is given by:

NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH

Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

The chemical reaction is given by:

NH_{3}+H_{2}O\rightarrow NH_4OH

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


8 0
3 years ago
What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




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3 years ago
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1 atom Mg, 2 atoms O and 2 atoms H.

1 + 2 + 2 = 5, so correct answer is C

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2 years ago
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