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Oduvanchick [21]

2 years ago

15

Two friends are discussing physical and chemical changes. Anna says that when you eat, it is a physical change because when you

chew, the food is only being changed into smaller pieces. Elsa says that when you eat, it is a chemical change because your body digests the food as the chemicals in your body break down the food into energy. Who is right?
A) Anna
B) Elsa
C) Both girls are correct
D) Neither girl is correct

2 answers:

Dmitrij [34]2 years ago

8 0

Answer:

It's B i just did the test. trust me

Kay [80]2 years ago

4 0

The person that was right about the chemical change is Elsa

Chemical changes can be regarded as one that takes place when chemical reactions occur and there is formation of new product such as chemical elements or compounds .

Physical change are one that doesn't result to formation of new chemical substances .

When someone eat because digestion occur due to breaking down of the food by enzymes into the small molecules the and as result of this new product is formed such as urine , energy and others

<em />

<em>Therefore, Elsa is right.</em>

Learn more at: brainly.com/question/1689737?referrer=searchResults

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a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

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Explanation:

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Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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