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astraxan [27]
3 years ago
15

You measure the emission spectrum of an atom that you suspect to be hydrogen. You have no absolute calibration of the wavelength

, but the ratios of the first 3 wavelengths in the series that you measure are 1 : 0.741 : 0.661. Which series is this (e.g. Lyman, Balmer, Paschen, Brackett)? Justify your answer.
Chemistry
1 answer:
trapecia [35]3 years ago
5 0

Answer:

Explanation:

What we need to do here is to determine the ratios by using  the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing  energy levels.

1/λ = Rh x ( 1/n₁² - 1/n₂²)

Lyman series  ( n₁=1 and n₂= 2,3 etc)  for the first two lines, the ratios will be:

1/λ₁ /1/λ₂  =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74  (the first ratio)

For Balmer series n₁ = 2  and n₂ = 3,4,5, etc

1/λ₁ /1/λ₂  =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)

Lets use the third line to check our answer:

1/λ₁ /1/λ₂  =(1/4 -1/3²) / (1/4 -1/5²) = 0.66

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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0
3 years ago
The options are: ( it can’t be repeated )
Artyom0805 [142]

Answer:

3- gamma radiation

Explanation:

Hello,

In the above question, 4 of the options are related to polymerization which are

1. Synthetic polymer

2. Natural polymer

3. Condensation polymerization

4. Addition polymerization.

The first two options are types of polymer that exists while the last two are polymerization techniques.

The odd option here which is "gamma radiation" is a particle which is emitted from radioactive substances during decay. It has no mass and no charge but it is highly penetrating and dangerous to human health.

However,

Synthetic polymers are also known as man made polymers and they exist around us because they're present in materials which we use everyday. An example is polyethylene, nylon-6,6 etc

Natural polymers are compounds which are polymeric in nature (compounds catenating to form a complex molecule). Natrual occurring polymers can be found in proteins and some lipids.

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Advantages and disadvantages of titration?
alisha [4.7K]

Answer:

What are the advantages of titration?

Titrimetric analysis commonly referred to as volumetric analysis offers distinct advantages over cumbersome gravimetric methods:

Speed of analysis.

Instantaneous completion of reactions.

Greater accuracy due to minimization of material loss involved in decanting, filtration, precipitation or similar operations.

Explanation:

Disadvantages

It is a destructive method often using up relatively large quantities of the substance being analysed.

It requires reactions to occur in a liquid phase, often the chemistry of interest will make this inappropriate.

It can produce significant amounts of chemical waste which has to be disposed of.

It has limited accuracy.

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