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ale4655 [162]
3 years ago
13

in the two stage cooling method what is the maximum amount of time allowed for both stages cooling food from 135 to 41 degrees F

Chemistry
2 answers:
dalvyx [7]3 years ago
5 0
<span>In the past, restaurants had four hours, straight through, to cool food to 41°F or lower. Now the FDA recommends cooling food in two stages -- from 135°F to 70°F in two hours then from 70°F to 41°F or lower in an additional four hours for a total cooling time of six hours.</span>
BlackZzzverrR [31]3 years ago
3 0
In the past, restaurants had four hours, straight through, to cool food to 41°F or lower. Now the FDA recommends cooling food in two stages -- from 135°F to 70°F in two hours then from 70°F to 41°F or lower in an additional four hours for a total cooling time of six hours
Explanation:
the two-stage cooling method<span> is a </span><span>Food Code </span>counselled<span> procedure for cooling food in restaurants and foodservice </span>institutions<span>. </span>within the<span> two-stage cooling </span>methodology<span>, food is</span><span> cooled from 140° F (60° C) to 70° F (21° C) </span>among 2<span> hours and to 41° F (5° C) or lower </span>among<span> four hours. Use of this cooling </span>methodology<span> ensures that food is cooled quickly and safely and has no harmful effects.</span>
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Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod
viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

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Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

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2 years ago
For every action, there is an equal and opposite reaction. true or false
Svetlanka [38]
True because evryone reacts a different way
5 0
2 years ago
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