The presence of lactic acid, produced during the lactic acid fermentation is responsible for the sour taste and for the improved microbiological stability and safety of the food. This lactic acid fermentation is responsible for the sour taste of dairy products such as cheese, yoghurt and kefir.
1.<span>Western world: yogurt, sourdough breads, sauerkraut, cucumber pickles and olives.
2.</span><span>Middle East: pickled vegetables.
3.</span><span>Korea: kimchi (fermented mixture of Chinese cabbage, radishes, red pepper, garlic and ginger)
4.</span><span>Russia: kefir.
5.</span><span>Egypt: laban rayab and laban zeer (fermented milks), kishk (fermented cereal and milk mixture)
6.</span><span>Nigeria: gari (fermented cassava)
7.</span><span>South Africa : magou (fermented maize porridge)
8.</span><span>Thailand : nham (fermented fresh pork)
9.</span><span>Philippines : balao balao (fermented rice and shrimp mixture)
</span><span>Lactic acid fermentation also gives the sour taste to fermented vegetables such as traditionally cultured sauerkraut and pickles. The sugars in the cabbage are converted into lactic acid and serve as a preservative.</span>
<h3>
Answer:</h3>
0.819 mol Ag
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
4.93 × 10²³ atoms Ag
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.818665 mol Ag ≈ 0.819 mol Ag
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
.5336
using avogadro’s law
