Answer:
semipermeability
Explanation:
partially but not freely or wholly permeable specifically : permeable to some usually small molecules but not to other usually larger particles a semipermeable membrane.
Answer:
The reaction of one mole of oxygen (O2) releases 445 kJ of energy.
Explanation:
Firstly, the reaction is exothermic since the sign of enthalpy change ΔH is negative.
The balanced equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): ΔH = −890 kJ,
Shows that 1 mole of CH₄ react with 2 moles of oxygen and releases 890 kJ.
So, every choice says that absorb is wrong (choice 1& 3).
Choice no. 4 is wrong since it says that 2 moles of methane releases 890 kJ, because only one mole release this amount of energy.
So, the right choice is The reaction of one mole of oxygen (O2) releases 445 kJ of energy.
Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
These are three questions and three complete answers
Answer:
a) Cr²⁺: [Ar] 4s² 3d²
b) Cu²⁺: [Ar] 4s² 3d⁷
c) Co³⁺: [Ar] 4s² 3d⁴
Explanation:
<u>a) Cr²⁺</u>
- Number of elecrons of the neutral atom: 24
- Number of electrons of the ion: 24 - charge = 24 - 2 = 22.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 22 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d²
<u>b) Cu²⁺</u>
- Number of elecrons of the neutral atom: 29
- Number of electrons of the ion: 29 - charge = 29 - 2 = 27.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 27 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d⁷
<u>c) Co³⁺</u>
- Number of elecrons of the neutral atom: 27
- Number of electrons of the ion: 27 - charge = 27 - 3 = 24.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 24 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d⁴
