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givi [52]
3 years ago
11

What are some milk products made from lactic acid fermentation?

Chemistry
1 answer:
Nadusha1986 [10]3 years ago
8 0
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>

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The paths in which electrons travel are called orbitals.

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How many mL of 0.013 M potassium hydroxide are required to reach the equivalence point in the titration of 75 mL 0.166 M hydrocy
umka21 [38]

Answer:

957.7mL

Explanation:

Using the formula below;

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Va = volume of acid (mL)

Cb = concentration of base (M)

Vb = volume of base (mL)

According to the information provided in this question:

Ca = 0.166 M

Cb = 0.013 M

Va = 75mL

Vb = ?

Using CaVa = CbVb

0.166 × 75 = 0.013 × Vb

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Vb =12.45/0.013

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2 years ago
Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
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Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

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Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

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2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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