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sattari [20]
2 years ago
6

What is negative in food sterilization?​

Chemistry
2 answers:
castortr0y [4]2 years ago
8 0

Answer:

Thermal food sterilization and pasteurization are the most widespread preservation technologies to extend food shelf life by inactivating microorganisms and enzymes that can deteriorate foodstuffs.

Airida [17]2 years ago
6 0

Explanation:

The process requires constant attention.

The equipment that it requires are costly.

Keeping moisture in food is difficult because of low moisture contents in the machine.

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PLEASE HURRY THIS IS TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
AURORKA [14]

Answer:

if i remember correctly i beleive its A  1.8 x 10^24

but im not for sure  also i think you forgot the 24

Explanation:

5 0
3 years ago
Noble gas notation for platinum
SOVA2 [1]

Pt

Hope this helps!!!

4 0
3 years ago
Calculate the average atomic mass of element X using the table below. Then use the periodic table to identify the element. Separ
Ksenya-84 [330]

Answer:

126.8, Iodine

Explanation:

  • mass ×abundance/100
  • (126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
  • 102.1+21.7+3=126.8

<em>IODINE</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomic</em><em> </em><em>mass</em><em> </em><em>of</em><em> </em>126.8 or 126.9

4 0
2 years ago
Draw all the structural isomers for the molecular formula C3H8O. Be careful not to draw any structures by crossing one line over
Art [367]

Answer:

The three isomers having the molecular formula C_{3} H_{8}O are drawn in the figure below.

Explanation:

5 0
2 years ago
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
2 years ago
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