Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles = 
= 
= 0.001245 moles
Concentration of KHP (C1) in litres = n/v
= 
= 0.062 mol/L
We know that:
= 
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>
Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
<h3>
Answer:</h3>
0.34 mol S
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
11 g S
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.343 mol S ≈ 0.34 mol S
thats kinda hot not even gon'
