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
I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
Answer:
Me and my friends were going to do a science experiment. Jonny’s job was to make the HYPOTHESIS. He said the “ If we mix baking soda and vinegar together, the TEMPERATURE will go down.”
So then Molly mixed the baking soda and vinegar together and checked the TEMPERATURE. We all OBSERVED as the thermometer’s TEMPERATURE went down. “ your THEORY/ HYPOTHESIS was correct!” Exclaimed Molly.
Then the whole science GROUP let out with a cheer! And wrote the information down on their EXPERIMENTAL info chart. They took a microscope and looked at the mixture because they wanted to the the little PARTICLES in the mixture. Lily CONTROLED the microscope she zoomed in and out to see the particles.
Explanation:
i hope this helps:)
Answer:
394.99g
Explanation:
The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)
n = nA ÷ 6.02 × 10^23
The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24
n = 2.45 × 10^24 ÷ 6.02 × 10^23
n = 2.45/6.02 × 10^(24-23)
n = 0.407 × 10¹
n = 4.07moles
Using mole = mass/molar mass
Where; molar mass of Fp3Bz2. is 97.05
g/mol.
mass = molar mass × mole
mass = 97.05 × 4.07
mass = 394.99g
