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andrezito [222]
3 years ago
8

How quickly must non-frozen ready-to-eat foods be consumed?

Chemistry
2 answers:
8090 [49]3 years ago
8 0

Answer:

24 hours

Explanation:

Frozen foods are foods that were put in a refrigerator. The essence of putting food in a refrigerator is to lower the rate at which microbes cause the spoilage of the food by lowering its temperature considerably.

After 24 hours, microbes must have commenced action on non-frozen ready to eat foods causing them to spoil in not more than 24hours.

Alexxandr [17]3 years ago
3 0

Answer:

i believe 24 hours

Explanation:

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What is the exact mass of one mole of carbon-12 atoms
love history [14]
One atom of carbon weighs exactly 12/6.022x10^23 = 1.9927x10^-23 grams<span>.</span>
3 0
4 years ago
Potassium permanganate as a symbol
Alexxx [7]

Answer: KMnO4-

Explanation:

You're looking at one potassium plus a polyatomic ion.

So K plus MnO4, equals:

KMnO4-

It also has a molar mass of 158.04 g/mol, I don't know if you need that, but I thought it would be nice to include it.

7 0
3 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
3 years ago
Can somebody plz help answer these true or false questions correctly thanks! (Only if u for sure know them) :)
asambeis [7]

Answer:

1. False

2. True

3. False

4. True

5. True

Explanation:

7 0
3 years ago
In a certain electrolysis experiment, 1.24 g of Ag were deposited in one cell (containing an aqueous AgNO3 solution), while 0.65
eduard

Answer:

The correct answer is 169.56 g/mol.

Explanation:

Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,

= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)

= 0.0115 mole of electron

The half cell reaction for the metal X is,  

X^3+ (aq) + 3e- = X (s)

From the reaction, it came out that 3 faraday will reduce one mole of X^3+.  

The molar mass of X will be,  

= 0.650 g/0.0115 *3 mol electron/1 mol

= 56.52 * 3

= 169.56 g/mol

7 0
3 years ago
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