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tino4ka555 [31]
3 years ago
11

Mention the advantage of Mercury over alcohol when they are used as thermal thermometric liquid​

Chemistry
1 answer:
kirill [66]3 years ago
6 0

Answer:

Alcohol has greater value of temperature coefficient of expansion than mercury.

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A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If
Ainat [17]

Answer:

V_2=52.2L

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L

Which is V1 in the Charles' law:

\frac{T_2}{V_2} =\frac{T_1}{V_1}

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K}  \\\\V_2=52.2L

Best regards!

8 0
3 years ago
When sodium is excited in a flame, two ultraviolet spectral lines at (wave length symbol)=372.1 nm and (wave length symbol)=376.
uranmaximum [27]
<span>Higher energy = shorter wavelength
Frequency is one cycle over an amount of time (seconds)
So higher frequency = higher energy = shorter wavelength</span>
5 0
4 years ago
Read 2 more answers
Calculate the molecular formula of a compound with the empirical formula C5H11 and a molar mass of 142.32 grams/mole
MA_775_DIABLO [31]
Relative formula mass C₅H₁₁ = 71

Now divide the molar mass by the RFM = 142.32 / 71 = 2

Now C₍₅ₓ₂₎H₍₁₁ₓ₂) = C₁₀H₂₂

Hope that helps
6 0
3 years ago
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Fynjy0 [20]

Answer:

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4 0
2 years ago
In the molecule ClF3, chlorine makes three covalent bonds. Therefore, three of its seven valence electrons need to be unpaired.
Kisachek [45]

Answer:

Explanation:

Chlorine has electronic configuration of 2 , 8 , 7

In n = 3 there are 7 electrons out of which 2 are in s , and 5 are in p . But out of 5 electrons in p , one electron jumps into d orbital . so the electronic configuration  becomes as follows

3s^23p_x^23p_y^23p_z^1  = 7

3s^23p_x^23p_y^13p_z^13d_{xy}^1

These orbitals like sp³d hybridise to form 7 degenerate orbitals out of which 2 orbitals contain electrons in pairs and rest three are singly occupied by electrons.( unpaired electrons )

3 0
3 years ago
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