Answer:
All of the above are true
Explanation:
a) The emission spectrum of a particular element is always the same and can be used to identify the element: It's true since the emission spectrum for each element is unique. It has the same bright lines at the same wavelength. This feature is used to identify elements. For example, the study of the emission spectra of light arriving from stars allow us to identify the elements presents in the star because the light contains the emission spectra of those elements.
b)The uncertainty principle states that we can never know both the exact location and speed of an electron: It is true since the velocity of an electron is related to its wave nature, while its position is related to its particle nature and we cannot simultaneously measure electron's position and velocity with precision.
c) An orbital is the volume in which we are most likely to find an electron: An orbital is a probability distribution map that is used to decribe the likely position of an electron in an atom.
Answer:
Mg2 + O2 → 2MgO
Explanation:
Hope this helps!! I got it right.
Answer:
101,37°C
Explanation:
Boiling point elevation is one of the colligative properties of matter. The formula is:
ΔT = kb×m <em>(1)</em>
Where:
ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-
kb is ebulloscopic constant (0,52°C/m)
And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):
203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>
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Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m
Replacing these values in (1):
X - 100°C = 0,52°C/m×2,64m
X - 100°C = 1,37°C
<em>X = 101,37°C</em>
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I hope it helps!
Explanation:
1 mol = 22.4 l
5.42 mol = 22.4 × 5.42 = 121.408
in two decimal place it is 121.41
1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.
2) So, one alkyl group is CH3 and second one can be CH or CH2.
3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.
4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.
The ratio remains the same, 3:2 ie 6:4
