Answer:
6.1×10^4Pa or 61KPa
Explanation:
The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.
Answer:
Fill in the blanks with the correct term.
a. a liquid that dissolves another substance.
b. a chemical that is dissolved.
c. a value used to describe the amount of one substance dissolved in another.
d. a liquid consisting of one substance dissolved in another.
Explanation:
a. A liquid that dissolves another substance is called the solvent.
b. A chemical that is dissolved solute.
c. A value used to describe the amount of one substance dissolved in another is called concentration.
d. A liquid consisting of one substance dissolved in another is called a solution.
It is given that the person weighs 62 kg = 62,000 g
Natural abundances in mass percent are:
O = 65%
C = 18%
H = 10%
N = 3.0%
Ca = 1.6%
P = 1.2%
Corresponding weights of the elements are:
O = 65/100 * 62000 g = 40.30 * 10^3 g
C = 18/100 * 62000 g = 11.16 * 10^3 g
H = 10/100 * 62000 g = 62.00 * 10^2 g
N = 3.0/100 * 62000 g = 18.60 * 10^2 g
Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g
P = 1.2/100 * 62000 g = 7.44 * 10^2 g
Answer:- There are 
moles.
Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is 
and it's molar mass is 294.31 grams per mole.
mg are converted to grams and then the grams are converted to moles as:
= moles of aspartame
So, there would be moles of aspartame in 1.00 mg of it.
% error = 3.4 %
Percent error = |accepted value - experimental value|/accepted value × 100%
∴ % error = |355 mL – 343 mL|/355 mL × 100 % = |12|/355 × 100 % = 3.4 %
