Answer:
<em>The electrons in an atom can only occupy certain allowed energy levels to a lower one</em>, the excess energy is emitted as a photon of light, with its wavelength dependent on the change in electron energy. This is why an atom can only emit specific wavelengths of light and not every possible wavelength.
Answer:
Mountain ranges are formed by a variety of geological processes, but most of the significant ones on Earth are the result of plate tectonics. Mountain ranges are also found on many planetary mass objects in the Solar System and are likely a feature of most terrestrial planets.
Answer:
<u>structural arrangements</u>
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<h2>properties of daimond: </h2><h3>appearance: transparent</h3><h3>hardness: very hard</h3><h3>thermal conductivity :very poor</h3><h3>electric conductivity: poor</h3><h3>density:</h3>

<h3>uses: jewellery and drilling</h3>
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<h2>properties of graphite:</h2>
<h3>appearance: black shiny</h3><h3>hardness: soft ,slippery to touch</h3><h3>thermal conductivity : moderate</h3><h3>electric conductivity: good</h3><h3>density:</h3>

<h3>uses:dry cell, electric arc, pencil lead, lubricant</h3>
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<h2>How Diamond and Graphite are chemically identical?</h2>
- On heating diamond or graphite in the air, they burn completely to form carbon dioxide.
- - Equal quantities of diamond and graphite when burned, produce exactly the same amount of carbon dioxide.
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<h2>Why the physical properties of diamond and graphite are so different?</h2>
Due to the difference in the arrangement of carbon atoms in diamond and graphite
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<h2>
<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>
Answer:
The answer to your question is P = 0.18 atm
Explanation:
Data
mass of O₂ = 0.29 g
Volume = 2.3 l
Pressure = ?
Temperature = 9°C
constant of ideal gases = 0.082 atm l/mol°K
Process
1.- Convert the mass of O₂ to moles
16 g of O₂ -------------------- 1 mol
0.29 g of O₂ ---------------- x
x = (0.29 x 1)/16
x = 0.29/16
x = 0.018 moles
2.- Convert the temperature to °K
Temperature = 9 + 273 = 282°K
3.- Use the ideal gas law ro find the answer
PV = nRT
-Solve for P
P = nRT/V
-Substitution
P = (0.018 x 0.082 x 282) / 2.3
-Simplification
P = 0.416/2.3
-Result
P = 0.18 atm