Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
Mitosis has 4 steps and meiosis has 5 steps so its 9 steps total here's a photo to help you if you need it
Answer:
It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.
Explanation:
The question is not very much clear.
If you are asking for molecules then 1 mole water= 6.023 * 10^23
If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3
If you are asking for particles then,
So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.
I hope that was helpful!
H=1 proton,1 electron
O=8 protons,8 neutrons and 8 electrons
total particles in one H2O molecule-28
total no. of particles in 1 mole of water- 6.023 * 10^23 * 28
453 divided by 224
density is roughly 2.02 g per ml
as a ml is 1 cm3 density is 2.02 grams per centimeter cubed
