Answer: Thus concentration of
in
is 0.011 and in
is 0.814
Explanation:
To calculate the concentration of
, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is 
We are given:

Putting values in above equation, we get:

The concentration in
is 
Thus concentration of
is
and 
Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar
Answer:
Its valancy is 1.
As it has the formula OH-.
It has valancy 1.
<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
Mass of oxygen = 2.2 g
Explanation:
Given data:
Mass of benzoic acid= 8.20 g
Mass of oxygen= ?
Solution:
Molar mass of oxygen = 16×2 g/mol
Molar mass of C₆H₅COOH = 7×12 + 1×6 + 2×16
Molar mass of C₆H₅COOH = 84 + 6 + 32
Molar mass of C₆H₅COOH = 122g/mol
Mass of oxygen in 8.20 g of C₆H₅COOH :
Mass of oxygen = 32 g.mol⁻¹/122 g.mol⁻¹ × 8.20 g
Mass of oxygen = 2.2g