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3 years ago

Explain how the sodium chloride works as a preservative

Chemistry

1 answer:

Olegator [25]3 years ago

8 0

Answer:

Eg:Salt...Essentially, water moves across a cell membrane to try to equalize the salinity or concentration of salt on both sides of the membrane. If you add enough salt, too much water will be removed from a cell for it to stay alive or reproduce. A high concentration of salt kills organisms that decay food and cause disease.

Explanation:

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In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is

Verizon [17]

Answer:

a) $10 mg NO_3^-/L$

b) $1.61*10^{-4}mol NO_3^-/L$

c) $2.26 mg N/L$

Carbon: $C=26.64 \frac{mg C}{L}$

Explanation:

<u>Nitrate</u>

First of all, is important to know that:

$1 ppm=1 mg/L$

a) 10 ppm of nitrate ( $NO_3^-$ ) is equal to $10 mg NO_3^-/L$

b) The molecular weight of nitrate is $62 g NO_3^-/mol$

$10 mg NO_3^-/L=0.01 g NO_3^-/L$

$\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L$

c) Nitrate has 14 mg of N per 62 mg of NO3

$10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L$

<u>Carbon</u>

Carbonate has 12 mg of C per 60 mg of $CO_3^{-2}$

Bicarbonate has 12 mg of C per 61 mg of $HCO_3^{-}$

$C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}$

$C=26.64 \frac{mg C}{L}$

8 0

3 years ago

If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti

gizmo_the_mogwai [7]

If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

Cb = 30/(68×500)

= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

= 0.00217m

Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

= 3.75

pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

learn more about pH:

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2 years ago

Write the correct name of the Fe(CIO3)3

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Answer:

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Explanation:

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defon

The correct answer is <span>Glucose and oxygen.</span>

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Answer:

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3 years ago

Explain how the sodium chloride works as a preservative​

Explain how the sodium chloride works as a preservative