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butalik [34]
2 years ago
15

Any entrée can be labeled "healthy" on a menu as long as it contains natural ingredients. True False

Chemistry
2 answers:
never [62]2 years ago
7 0
I'm pretty sure this statement is False.
bagirrra123 [75]2 years ago
3 0

Answer:

False.

Explanation:

Entrée is a culinary term used for a course during a meal service. In parts of the US and Canada, the word entrée is often used to refer to the main dish that is served while in other parts of the world, it usually means the starter or the simple meal that is served before the main dish / course. Entrées can be made of any component but the content of the natural ingredients does not define its healthy status. Thus, it is false to say that any entrée can be labelled "healthy" on a menu as long as it contains natural ingredients.

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Determine the number of moles of water assicated with the salt.
Nookie1986 [14]

Answer:

Divide the mass of your anhydrous (heated) salt sample by the molar mass of the anhydrous compound to get the number of moles of compound present. In our example, 16 grams / 160 grams per mole = 0.1 moles. Divide the mass of water lost when you heated the salt by the molar mass of water, roughly 18 grams per mole.In order to determine the formula of the hydrate, [Anhydrous Solid⋅xH2O], the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid (Equation 2.12. 6).

6 0
2 years ago
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

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