Answer:
The answers are in the explanation
Explanation:
- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.
- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.
- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.
- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.
I hope it helps!
The functional group can affect the chemical properties and physical properties. Like CH3CH3 and CH3CH2OH. The first one is gas and second one is liquid. The first one is not easy to react with others while the second one is.
No of moles = given mass ÷ molecular mass
n = 55.98 ÷ (12+19×2+35.5×2)
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.