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3 years ago

Bob adds heat to a liquid substance. When enough heat is added,

1 answer:

8 0

Bob adds heat to a liquid substance. When enough heat is added, the liquid would undergo a phase transition from the liquid to the vapor phase. This process is called evaporation. It is an endothermic process where heat should be supplied to the system for the process to happen. The temperature of the system, when enough heat is supplied to allow phase transition, is called the boiling point. It is the temperature when the substance starts to boil in turn to a gas or a vapor. The heat that is associated during boiling is called the latent heat of vaporization. It is the heat that is absorbed during the phase change without changing the temperature of the system.

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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

In Environment A, trees are growing back and some of the larger animals are returning to a logged forest. In Environment B, lich

laiz [17]

Answer:

Environment A is not undergoing succession, and Environment B is.

Explanation:

Ecological succession is a gradual process in which ecosystems significantly change over time. Ecological succession is a term used by scientists to describe the change in the structure of a community of different species, or ecosystem. This concept of ecological succession stems from a desire to understand the patterns of change in large and complex ecosystems like forests and how they can exist in places known to be recently formed, such as volcanic islands.

In environment A, the ecosystem is not really changing, organisms are merely returning to their natural habitat. It does not represent any change in the ecosystem.

In environment B, the original ecosystem has become grossly modified, first by the appearance of lichen and mosses and subsequently by grasses shrubs and animals. These sequence of events correlate well with the idea of ecological succession presented in the opening paragraph hence environment B is undergoing ecological succession.

4 0

3 years ago

Read 2 more answers

What is it called when particles travel from an area of high consentration to an area of low consentration?

Umnica [9.8K]

If it isn't using energy, it's called diffusion

8 0

3 years ago

Potassium impart purple colour but beryllium do not impart any colour to the flame.Why?

BaLLatris [955]

Answer:

<h2>The electrons in beryllium and magnesium are too strongly bound to get excited by flame. Hence, these elements do not impart any color.</h2>