Answer:
A.) Brønsted-Lowry bases
Explanation:
Amines have a lone pair of electrons.
Brønsted-Lowry bases donate a lone pair of electrons in exchange for a hydrogen ion.
Therefore, if exposed to an acid, amines will give up electrons in order to bond with a hydrogen. This makes them Brønsted-Lowry bases.
5.0x10^1 kg is the correct answer
Answer:
The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Explanation:
Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:
CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O
The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio
Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M
Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles
Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles
Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5
There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.
Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.
Answer:
V₂ = 111.3 mL
Explanation:
Given data:
Initial volume of gas = 50.0 mL
Initial temperature = standard = 273.15 K
Final volume = ?
Final temperature = 335 °C (335+273.15 = 608.15 K)
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 50.0 mL ×608.15 K / 273.15 k
V₂ = 30407.5 mL.K / 273.15 K
V₂ = 111.3 mL
