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tensa zangetsu [6.8K]
3 years ago
5

Is cyclohexanol an alkane, alkene, or an alcohol?

Chemistry
1 answer:
Oxana [17]3 years ago
8 0
Cyclohexanol is an Alcohol..........
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Which of the following is a possible set of quantum numbers for an electron (n,l,m0,ms) ?
Mashutka [201]

Answer is: (3, 2, 0, -1/2).

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.

For principal quantum number n=3:  

1) azimuthal quantum number (l) can be l = 0...n-1:  

l = 0, 1, 2.  

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.  

2) magnetic quantum number (ml) can be ml = -l...+l.  

ml = -2, -1, 0, +1, +2.

Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.  

3) the spin quantum number (ms), is the spin of the electron.  

ms = +1/2, -1/2.  

(1, 1, 0, +1/2)  is not correct because orbital quantum number cannot be l = 1 for n = 1.

(2, 1, 2, +1/2)  is not correct because magnetic quantum number cannot be ml = 2 for orbital quantum number l = 1.

(3, -2, 1, -1/2) is not correct because orbital quantum number cannot be l = -2 for principal quantum number n = 3.

5 0
3 years ago
The normal boiling point of a liquid is 282 °C. At what temperature (in
ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

ln\left (\frac{p_{2}}{p_{1}}  \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}}  \right )

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

\Delta H_{vap} = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

ln\left (\frac{0.2}{1}  \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15}  \right )

\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0
3 years ago
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