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Anit [1.1K]
3 years ago
8

Kimchi is traditional Korean food made by fermenting cabbage or other vegetables in a mixture of spices and salt. Kimchi origina

lly developed as a method of preserving vegetables before the advent of modern refrigeration. Why does the use of a salt solution in Kimchi act as a preservative against bacterial decay?
A. Water passes out of the cabbage leaves into the salt solution, wilting the leaves.
B. Water passes out of the salt solution and into the bacteria, lysing the bacterial cells.
C. Water passes into the salt solution, dehydrating bacterial cells and making them harmless.
D. The salt solution used to preserve kimchi has a lower molality than that inside most bacteria.
Chemistry
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

C. Water passes into the salt solution, dehydrating bacterial cells and making them harmless.

Explanation:

The salt solution is hypertonic to the bacterial cells and as such, water molecules will move from the bacterial cells into the salt solution, dehydrating the cells and rendering them harmless.

Option A is also true but it is irrelevant to the question asked. Option B and D are wrong.

The correct option is C.

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23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
3 years ago
Draw the major product formed when the structure shown below undergoes substitution in ch3ch2oh with heat. (use a wavy bond (sin
muminat

Answer:

Explanation:

The missing image is attached below.

The objective of this question is to draw the major product formed from the diagram attached below.

From the diagram attached, we will see the reaction of a tertiary alkyl halide together with a weak nucleophile (ch3ch2oh) undergoing a nucleophilic substitution (SN₁) mechanism to yield a racemic mixture(i.e., compound that is not optically active but contains an equal amount of dextrorotatory and levorotatory stereoisomers) as a product.

7 0
3 years ago
4. How would you calculate the poH of a base when given the<br><br> concentration of [OH]-?
Novay_Z [31]

Answer:

Use pOH = -log₁₀ [OH-]

Explanation:

pOH can be calculated from the concentration of hydroxide ions using the formular below:

pOH = -log₁₀ [OH-]

The pOH is the negative logarithm of the hydroxide ion concentration.

3 0
3 years ago
How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
Answers fast please Gets oxygen into your body and carbon dioxide out
yKpoI14uk [10]
B respiratory is correct
4 0
3 years ago
Read 2 more answers
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