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Tom [10]
3 years ago
5

One source of aluminum metal is alumina, Al2O3. a. Determine the percentage composition of Al in alumina. b. How many pounds of

aluminum can be extracted from 2.0 tons of alumina.?
Chemistry
1 answer:
Gnom [1K]3 years ago
3 0

Answer:

sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer

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Why chlorine is more reactive than hydrogen​
Lynna [10]

Answer:

<em>Hi Todoroki here!!! </em>

Explanation:

Chlorine has the electron configuration [Ne]3s 2 3p 5, with the seven electrons in the third and outermost shell acting as its valence electrons. Like all halogens, it is thus one electron short of a full octet, and is hence a strong oxidising agent, reacting with many elements in order to complete its outer shell.

<em>Your welcome!!</em>

7 0
3 years ago
Define a transverse wave
Mumz [18]
Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
4 0
3 years ago
Which substance can be broken down by a chemical change
ELEN [110]
Lots of substances. Rocks, metals, minerals, you name it.
5 0
3 years ago
Read 2 more answers
Help please. 15.0 moles of gas are in a 3.00L tank at 23.4∘C . Calculate the difference in pressure between methane and an ideal
CaHeK987 [17]
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT.  Substituting  a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
3 0
3 years ago
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
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