Since the molecules of a gas are constantly in motion, moving and hitting other gas molecules in all directions of the enclosed container, the molecules transfer a portion of energy through the interaction of other gas particles as they hit, but then bounce off.
Answer:
The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode
Reaction:
<span>HCl + NaOH ---> NaCl + H2O
</span><span>1 mole of HCl = 36,5 g
</span><span>1 mole of NaOH = 40g
</span><span>so, according to the reaction:
</span><span>1 mol HCl = 1 mol NaOH
</span>so, we need > 36,5 g HCl (<u>hydrochloric acid</u><span>)
</span><u>
answer: 36,5 g HCl (hydrochloric acid)
</u><span> ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
</span><span>next question.
</span><span>
1 mole of NaCl = 58,5 g
</span><span>1 mole of H2O = 18g
</span>
so, according to the reaction:
1 mole of HCl (36,5 g) <span>----------------- - 1 mole of NaCl (58,5 g)
</span><span>(the same for NaOH)
i
</span>1 mole of HCl<span> (36,5 g) ------------------ 1 mole of H2O (18 g)
</span>(the same for NaOH)
<span>so, this reaction is stechiometric
</span><u>
answer: 58,5 g NaCl i 18g H2O</u>
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
There are 7 signifigant figures in this number!
