Calculate h30+ for a 8.86*10^3 hbr solution
1 answer:
[H₃O⁺] = 8.86 x 10³ M
<h3>Further explanation</h3>The dissociation of water :
H₂O(l)⇒H⁺+OH⁻
Hydrogen ions H⁺ bond with H₂O to form H₃O⁺ ions(hydronium ion)
When acid(Hbr acid) is in the presence of water(dissociates), the H⁺ ions bond with water molecules to form hydronium, the reaction :
HBr(aq)+H₂O→H₃O+(aq)+Br−(aq
So Ionization of HBr :
[HBr]= 8.86 x 10³,
From equation ratio [H₃O⁺] : HBr = 1 : 1, so
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<h3>Answer:</h3>
3.3 × 10²³ molecules Cu(NO₃)₂
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
0.55 mol Cu(NO₃)₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂
Answer:
Fe =52.2%, O = 44.9%, H = 2.81%
Explanation:
Percentage composition is also known as percentage by mass.
First, find the Mr of the compound.
Mr of Fe(OH)₃ = 55.8 + (16 × 3) + (1 × 3)
= 106.8
Now Divide the mass of the element, as per the compound, by the Mr of the compound found and times that by 100.
% composition of Fe = × 100 = 52.2%
% composition of O = × 100 = 44.9%
[There are 3 oxygen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of oxygen with the number of atoms in the compound: 16×3 ]
% composition of H = × 100 = 2.81%
[There are 3 hydrogen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of hydrogen with the number of atoms in the compound: 1×3 ]