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>
Answer:For a typical experiment, you should plan to repeat it at least three times (more is better).
Explanation:
Answer:
Explanation:
Polarity is about differencens in electronegativity. CH bonds have around the same electronegativity value so a CH bond is nonpolar. The more CH bonds there are in a molecule, the more nonpolar it is. Since CH3CH2OH has more carbon-hydrogen bonds than CH3OH, it is more nonpolar. With the same reasoning, since CH3OH has less CH bonds, it's more polar.
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
![Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D%7D%7B%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D%7D)
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both ![[\mathrm{SO_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BSO_2%5C%2C%20%28g%29%7D%5D)
and ![[\mathrm{O_2\, (g)}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BO_2%5C%2C%20%28g%29%7D%5D)
will increase if the pressure is increased through compression. However, because 
and 
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient 
.
As a result, the increase in pressure will have no impact on the value of 
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
The molecular mass of Carvone is calculated as;
= 12 (C)₁₀ + 1.008 (H)₁₄ + 16 (O)
= 120 + 14.112 + 16
= 150.112
%age of Carbon;
= (120 ÷ 150.112) × 100
= 79.94 %
%age of Hydrogen;
= (14.112 ÷ 150.112) × 100
= 9.40 %
%age of Oxygen;
= (16 ÷ 150.112) × 100
= 10.65 %
