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3 years ago

14

In which situation is chemical energy being converted to another form of energy? a lamp plugged into the electric grid a flutter

ing flag a floating wooden log a burning candle

1 answer:

Irina18 [472]3 years ago

5 0

OK so the answer to your question will be option D. because chemical energy is being transferred as heat and light

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The strongest elemental oxidizing agents are found in the _________ _________ region of the periodic table while the strongest r

Slav-nsk [51]

Strongest reducing agents are in Group 1 . For example lithium. The strongest oxidising agents are in Group 7 , For example Fluorine.

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3 years ago

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Which of the following is not found in the nucleus of an atom?

DochEvi [55]

Electron - In an atom, the central part which is composed of neutrons and protons including quarks is called the nucleus. Whereas the electron is the particle with negative elementary electric charge can be found inside the atom surrounding the nucleus.

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3 years ago

Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

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3 years ago

A sprinter set a high school record in track and field, running 200.0 m in 21.7 s . What is the average speed of the sprinter in

Yuliya22 [10]

Given data:

Distance travelled (d)= 200.0 m

Time taken (t) = 21.7 s

To determine:

The average speed in km/hr

Calculation:

Convert distance from m to km

1000 m = 1 km

200.0 m = 1 km * 200.0 m/1000 m = 0.2 km

Convert time from sec to hours

3600 sec = 1 hour

21.7 sec = 1 hour * 21.7 sec/3600 sec = 6.027 *10^-3 hrs

Speed is defined as the amount of distance travelled per unit time

Speed = distance/time = 0.2 km/6.027*10^-3 hr

= 33.18 km/hr (or 33.2 km/hr)

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2 years ago

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Ostrovityanka [42]

Electronic items contain lcds

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2 years ago

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