Answer:
SO₃(g) + H₂O(l) → H₂SO₄(aq)
Explanation:
The<em> molecular formula for the involved species</em> are:
- Sulfur trioxide = SO₃. ("trioxide" indicates the presence of 3 oxygen atoms).
With the above information in mind we can proceed to write the reaction equation:
- SO₃(g) + H₂O(l) → H₂SO₄(aq)
Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;
where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;
Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
Answer:
a substance made by mixing other substances together.
Explanation:
A mixture is a substance made by mixing two or more substances together.
The anion<span> is also </span>larger than<span> the </span>atom<span> because of </span>electron-electron repulsion<span>. As more </span>electrons are<span> added to the </span>outer shell<span>, and even to </span>higher<span> principle energy levels, the </span>repulsion<span> bewteen the negatively charged particles grows, pushing the </span>shells<span> farther from the nucleus.</span>
Hydroxylamine in water: HONH₂(aq) + H₂O(l) ⇄ HONH₃⁺(aq) + OH⁻(aq).
Hydroxylammonium nitrate in water: HONH₃NO₃(aq) → OHNH₃⁺(aq) + NO₃⁻(aq).
1) with positive hydrogen ions (protons) react base and gives weak conjugate acid:
H⁺(aq) + HONH₂(aq) ⇄ HONH₃⁺(aq).
2) with hydroxide anions react acid and produce weak base and weak electrolyte water:
HONH₃⁺(aq) + OH⁻(aq) ⇄ HONH₂(aq) + H₂O(l).
