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drek231 [11]
3 years ago
11

Why does hair take longer to dry after a shower on days with high relative humidity

Chemistry
1 answer:
Luden [163]3 years ago
3 0
After you take a shower your hair is still wet. When there is humidity there is more water vapor in the air. So the air around you carries more water. If there is little to NO humidity your hair dries faster because there is less water vapor in the air and the wetness of your hair can escape faster in the form of water vapor. But in humidity the water from your hair needs to "fit into the air" which takes longer because of the WV already in the air. I hope this helped:-)!
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Is chromate polar or nonpolar?
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Answer:

i THINK so

Explanation:

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Construct a simulated proton-decoupled 13C NMR spectrum for tert-butyl alcohol. Drag the resonance signal icon to the appropriat
atroni [7]

Answer:

                     The ¹³C-NMR Spectrum of  <em>tert</em>-butyl alcohol will show only two signals.

(i) Signal at around 31 ppm:

                                              This signal towards upfield is for the carbon atoms which are more shielded and are having rich electron surroundings. The height of peak at y-axis shows the number of carbon atoms as compared to other peaks. In this case it is three times the height of second signal hence, it shows that this peak corresponds to three carbon atoms.

(ii) Signal at around 70 ppm:

                                              This signal towards downfield is for the carbon atom which is more deshielded and is having electron deficient surrounding. As compared to the second signal the height of this peaks corresponds to only one carbon. And the deshielded environment shows that this carbon is directly attached to an electronegative element.

4 0
4 years ago
Calculate the total energy, in kilojoules, that is needed to turn a 46 g block
gogolik [260]

<u>Answer:</u> The amount of heat absorbed is 141.004 kJ.

<u>Explanation:</u>

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1): H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)

(2): H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)  

(3): H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)

(4): H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)

Calculating the heat absorbed for the process having the same temperature:

q=m\times \Delta H_{(f , v)}       ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and \Delta H_{(f , v)} is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

q=m\times C_{s,l}\times (T_2-T_1)      ......(ii)

where,

C_{s,l} = specific heat of solid or liquid

T_2\text{ and }T_1 are final and initial temperatures respectively

  • <u>For process 1:</u>

We are given:

m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC

Putting values in equation (i), we get:

q_1=46g\times 2.108J/g^oC\times (0-(-25))\\\\q_1=2424.2J

  • <u>For process 2:</u>

We are given:

m=46g\\\Delta H_{fusion}=334J/g

Putting values in equation (i), we get:

q_2=46g\times 334J/g\\\\q_2=15364J

  • <u>For process 3:</u>

We are given:

m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC

Putting values in equation (i), we get:

q_3=46g\times 4.186J/g^oC\times (100-0)\\\\q_3=19255.6J

  • <u>For process 4:</u>

We are given:

m=46g\\\Delta H_{vap}=2260J/g

Putting values in equation (i), we get:

q_4=46g\times 2260J/g\\\\q_4=103960J

Calculating the total amount of heat released:

Q=q_1+q_2+q_3+q_4

Q=[(2424.2)+(15364)+(19255.6)+(103960)]

Q=141003.8J=141.004kJ                  (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.

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