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kati45 [8]
3 years ago
10

Which of the following is not a function?

Chemistry
1 answer:
EastWind [94]3 years ago
8 0

Answer:

A student's name paired with the sport that they play.​

Explanation:

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Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transit
Lapatulllka [165]

<u>Answer:</u> The wavelength of spectral line is 656 nm

<u>Explanation:</u>

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.097\times 10^7m^{-1}

n_f = Final energy level = 2

n_i= Initial energy level = 3

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.524\times 10^6m^{-1}}=6.56\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.56\times 10^{-7}m\times (\frac{10^9nm}{1m})=656nm

Hence, the wavelength of spectral line is 656 nm

6 0
3 years ago
Middle School science 3 1. Illustrate the position of the Earth, moon, and sun during a lunar eclipse. You may do so by typing a
lilavasa [31]
I hope this helps you

7 0
3 years ago
The melting point of aluminum is 660ºC. At what temperature will aluminum melt if you have 5 grams? 10 grams? Explain your answe
valkas [14]

Answer:

the melting point of aluminum is 660 degrees Celsius.

5 0
2 years ago
What happens to water during evaporation?
kotegsom [21]
D would be your best bet because evaporation occurs when water is heated, it then vibrates and then magic!
5 0
3 years ago
Read 2 more answers
How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
bezimeni [28]

Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

6 0
3 years ago
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