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butalik [34]
3 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]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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Static electricity is the buildup of electric charges in
Sauron [17]
Static electricity is the buildup of electric charges on surface of an object.
static electrical charges remain on the surface of the object until they bleed off to the ground or until they get quickly neutralised by a discharge.
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4 years ago
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4 years ago
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I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
Please help me with this homework
aleksandr82 [10.1K]
Answer is gas for sure and I think it’s ice for question 2
5 0
3 years ago
Each of the chemically active Period 2 elements forms stable compounds in which it has bonds to fluorine.(a) What are the names
Softa [21]

The names of the compound are lithium fluoride ( LiF) , beryllium difluoride (BeF2) , Boron trifluoride (BF3) , carbon tetrafluoride (CF4) , nitrogen trifluoride (NF3) , oxygen difluoride (OF2) .

Lithium is the first element of period 2 which reacts with fluorine to form LiF ( lithium fluoride ) . it is an inorganic compound . it is also a colorless solid . it is less soluble in water . it is chemically stable because of its comparable molecular mass .

Beryllium is the second element of period 2 which reacts with fluorine to give beryllium difluoride (BeF2) . it is inorganic compound . it is highly soluble in water. it is also a stable compound . it have low melting point .

Boron is the third element of period 2 which reacts with fluorine to form

BF3 (Boron trifluoride ) . it is a inorganic compound . it is colorless and toxic gas forms  . it is stable in dry atmosphere but its octet is not satisfied .

Carbon is the 4th element of the period 2 which reacts with fluorine to form carbon tetrafluoride (CF4) . it is not soluble in water . it is a greenhouse gas . it dissolves in oil. it is very stable compound .it forms covalent bond .

Nitrogen is the 5th element of period 2 which reacts with fluorine to form nitrogen trifluoride (NF3) . it is also a inorganic compound . it  is colorless and non-flammable .  it is a stable gas at room temperature .

Oxygen is the 6th element of period 2 which reacts with fluorine to form oxygen difluoride (OF2) . it is colorless poisonous gas . it is partially stable or relatively stable .

Neon is a noble gas and also a stable element . it is odorless and colorless . so it is nonreactive . so it doesn't form bond with fluorine .

<h3>Learn more about Fluorine here :</h3>

brainly.com/question/3494441

#SPJ4

8 0
2 years ago
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