Answer is: f<span>ormula for the hydrated compound is CuSO</span>₄·3H₂O.
ω(H₂O) = 25,3% = 0,253.
ω(CuSO₄) = 100% - 25,3%.
ω(CuSO₄) = 74,7% = 0,747.
ω(H₂O) : M(H₂O) = ω(CuSO₄) : M(CuSO₄).
0,253 : M(H₂O) = 0,747 : 159,6 g/mol.
M(H₂O) = (0,253 · 159,6 g/mol) ÷ 0,747.
M(H₂O) = 54 g/mol.
N(H₂O) = 54 g/mol ÷ 18 g/mol.
N(H₂O) = 3.
Answer:
In simple words, in an open system, the system that has external interactions. Such interactions can take the form of information, energy, mass etc. So keeping this definition in mind, the law of conservation of mass in open system is difficult because when chemical reaction takes places things change form and some things are released as energy, That energy than escapes the system and interacts with other surrounding. This is why it is difficult to demonstrate Law of Conservation of Mass because the changes in mass due to chemical reaction cannot be accounted for properly since the lost mass (in form of energy) was lost to environment.
Whereas, in closed system, all components occur inside a closed container or something where external interactions are minimal to none so the energy lost due to chemical reaction does not escape the system thus it can be accounted for properly to demonstrate the law of conservation of Mass.
Explanation:
Too many words but should get the point across hopefully!
The act of using senses or tools to gather information is making an observation
Answer:
a) pH = 13.176
b) pH = 13
c) pH = 12.574
d) pH = 7.0
e) pH = 1.46
f) pH = 1.21
Explanation:
HBr + NaOH ↔ NaBr + H2O
∴ equivalent point:
⇒ mol acid = mol base
⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)
⇒ Va = 0.025 L
a) before addition acid:
⇒ <em>C </em>NaOH = 0.150 M
⇒ [ OH- ] = 0.150 M
⇒ pOH = - Log ( 0.150 )
⇒ pOH = 0.824
⇒ pH = 14 - pOH
⇒ pH = 13.176
b) after addition 5mL HBr:
⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M
⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M
⇒ [ OH- ] = 0.1 M
⇒ pOH = 1
⇒ pH = 13
c) after addition 15mL HBr:
⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M
⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M
⇒ [ OH- ] = 0.0375 M
⇒ pOH = 1.426
⇒ pH = 12.574
d) after addition 25mL HBr:
equivalent point:
⇒ [ OH- ] = [ H3O+ ]
⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²
⇒ [ H3O+ ] = 1 E-7
⇒ pH = 7.0
d) after addition 40mL HBr:
⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M
⇒ [ H3O+ ] = 0.035 M
⇒ pH = 1.46
d) after addition 60mL HBr:
⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M
⇒ [ H3O+ ] = 0.062 M
⇒ pH = 1.21
