By lowering temperature from liquid to solid, atoms lose some of their energy, when atoms slow down, they move close together and make the matter denser.
The required solution is prepared taking 0.1 mL of 10 M stock NaOH solution and placing it into a 100 mL volumetric flask. The solution is made up to 100 mL with distilled water in order to have a 100 mL solution of 10 mM NaOH solution.
The required solution is prepared by diluting a given volume of the stock solution.
Using the dilution formula: C₁V₁ = C₂V₂
Where C₁ = initial concentration of solution
V₁ = initial volume of solution
C₂ = final concentration of solution
V₂ = final volume of solution
From the given values:
C₁ = 10 M, V₁ = ?, C₂ = 10 mM = 0.01 M, V₂ = 100ml
Making V₁ subject of the formula in the dilution formula:
V₁ = C₂V₂/C₁
V₁ = 0.01 * 100 / 10
V₁ = 0.1 mL
Therefore, the volume of stock solution needed to prepare 100 mL of 10 mM NaOH solution is 0.1 mL.
0.1 mL of 10 M stock NaOH solution is taken and put into a 100 mL volumetric flask. The solution is made up to 100 mL with distilled water in order to have a 100 mL solution of 10 mM NaOH solution.
Learn more more about preparation of solutions and dilution at : brainly.com/question/24549357
Answer:
physcial change
Explanation:
As it has visible changes on shape and size
Nitrogen — 78 percent.
Oxygen — 21 percent.
Argon — 0.93 percent.
Carbon dioxide — 0.04 percent.
Trace amounts of neon, helium, methane, krypton and hydrogen, as well as water vapor.
Answer:
43.47 g
Explanation:
The <em>boiling point elevation</em> is described as:
Where ΔT is the difference in boiling points: 120.6-118.4 = 2.2 °C
K is the boiling point elevation constant, K= 2.40 °C·kg·mol⁻¹
and m is the molality of the solution (molality = mol solute/kg solvent).
So first we <u>calculate the molality of the solution</u>:
- 2.2 °C = 2.40 °C·kg·mol⁻¹ * m
Now we calculate the moles of benzamide (C₇H₇NO, MW=315g/mol), using the given mass of the liquid X.
- 150 g ⇒ 150/1000 = 0.150 kg
- 0.917 m = molC₇H₇NO / 0.150kg
Finally we convert moles of C₇H₇NO into grams, using its molecular weight:
- 0.138 molC₇H₇NO * 315g/mol = 43.47 g
