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ira [324]
3 years ago
9

Explain how the understanding of atomic emission spectrum led to development of the atomic theory? ( 4 points)

Chemistry
2 answers:
dangina [55]3 years ago
6 0
The atomic emission spectrum (Bhor analyzed) of hydogren and showed mathematically that the result of movement making a transition from a high energy state to a lower energy.

Hope this help
-Dominant- [34]3 years ago
5 0
Well, all of this we owe it to Bohr who analyzed the atomic emission spectrum of hydrogen and he could probe matematically that it was a result of movement of e- from an especific energy level to a lower one. The understanding of levels of energy took to the development of the atomic theory
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about 82.2 centimetres.
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Barium and lead both produce yellow precipitate with the chromate ion as part of their confirmation tests. if you had a sample t
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Barium is an element that is primarily used in fireworks because of its distinct green colour while on the other hand, lead is used in electrical industries due to its unique properties. The ions of these elements can simply be identified when its insoluble salts get precipitated by any process.
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3 years ago
Read 2 more answers
Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,
ludmilkaskok [199]

<u>Answer:</u> The empirical formula for the given compound is H_{4}O_1N_1Cl_1=H_4NOCl

<u>Explanation:</u>

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles

Moles of Chlorine = \frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = \frac{5.80}{1.44}=4.03\approx 4

For Oxygen = \frac{1.44}{1.44}=1

For Nitrogen = \frac{1.44}{1.44}=1

For Chlorine = \frac{1.44}{1.44}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is H_{4}O_1N_1Cl_1=H_4NOCl

3 0
3 years ago
What is the name of this molecule?
Ludmilka [50]
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6 0
3 years ago
Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.
Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n}{W_s}

where,

n = moles of solute

W_s = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg

Now put all the given values in the formula of molality, we get

Molality=\frac{0.241}{0.0137kg}=17.6mole/kg

Therefore, the molality of solution is 17.6 mole/kg

3 0
3 years ago
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