<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
Answer:
2.74 M
Explanation:
Given data:
Mass of sodium chloride = 80.0 g
Volume of water = 500.0 mL
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will convert the mL into L.
500.0 mL× 1 L /1000 mL = 0.5 L
In next step we will calculate the number of moles of sodium chloride.
Number of moles = mass/molar mass
Number of moles = 80.0 g/ 58.4 g/mol
Number of moles = 1.37 mol
Molarity:
M = 1.37 mol/ 0.5 L
M = 2.74 M
75% i think it is consumers
25% i think it is producer
Answer:
5.702 mol K₂SO₄
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Compounds
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 993.6 g K₂SO₄
[Solve] moles K₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of K: 39.10 g/mol
[PT] Molar Mass of S: 32.07 g/mol
[PT] Molar mass of O: 16.00 g/mol
Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄