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3 years ago

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

Chemistry

2 answers:

never [62]3 years ago

7 0

I'm pretty sure this statement is False.

bagirrra123 [75]3 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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In a neutralization reaction, an aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to p

dem82 [27]

An aqueous solution of an arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt.

This is a compound which is formed as a result of a neutralization reaction between acid and base.

Arrhenius acid reacts with an aqueous solution of an arrhenius base to produce water and salt due to increased concentration of H+ and OH- respectively.

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2 years ago

Which of the following is not a correct chemical equation for a double displacement reaction? (D?)

emmasim [6.3K]

The correct answer would be the last option. A double displacement type of reaction involves the switching of places the cations and anions accordingly. The given reaction is erroneous since in the product side the anions and cations are being paired which would not make sense. The correct reaction should be

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3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,