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KIM [24]
3 years ago
13

Why is ice less dense than liquid water?

Chemistry
2 answers:
podryga [215]3 years ago
8 0
Hey there,

Your question states: Why is ice less dense than liquid water?
There reason why Ice is less dense than liquid is because ice has compact molecules that are tight together and they stay still. But as for water, those molecules are lose and they are more free than ice.

Hope this helps.
~Jurgen
rosijanka [135]3 years ago
6 0

The answer is: Ice expands (volume is increasing) when it freezes, because of hydrogen bonds.

When molecule frezes, it lose energy. When molecules are far apart, it means the volume is greater and it expands.

For example, ice expands when water is freezing.

Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.

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5x^2 + 3×^2-x+7x<br>Help plz​
PSYCHO15rus [73]

Answer:

5x^2 + 3x^2 -x +7x

5x^2+3x^2+6x = 8x^2+6x

factorize it to get

2x(4x+3)

3 0
3 years ago
Let’s say that you have a solvatochromic compound that appears red in a solvent. You dissolve the compound in another solvent an
EastWind [94]

Answer:

Hypsochromic compound, More polar solvent

Explanation:

Hypsochromic shift refers to the shift of solution colour to blue side of the visible spectrum (blueshift) with increasing polarity of the solvent. In our case, the solution changes to orange colour from red when solvent is changed. This means that the emission spectrum of the solution underwent blueshift. (As orange colour is on the 'blue' side for red colour.) So this is a hypsochromic shift, and the new solvent is more polar that the previous one, as it caused hypsochromic shift.

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3 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

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