All categories

3 years ago

In what ways is the process of digestion a physical change?

2 answers:

7 0

When food is physically changed, mechanical digestion occurs. Food is broken into smaller parts and mixes continually with enzymes and other gastric juices. Mechanical digestion occurs in the mouth, stomach, and small intestine. Food is chemically changed in digestion when new, smaller substances are formed.

Mekhanik [1.2K]3 years ago

3 0

Answer:

When the teeth chews the food. Food moves down the esophagus by the process of peristalsis .The stomach churns the food which breaks it into smaller pieces and mixes it with the gastric juices to produce a liquid called chyme. In the small intestine a process called segmentation the chyme sloshes back and forth between the segments of the small intestine.

Explanation:

You might be interested in

What type of particles move to create electricity

yKpoI14uk [10]

Answer:

Electrons

Explanation:

6 0

3 years ago

Read 2 more answers

What is reaction rate

zimovet [89]

Answer:The change in concentration of a reactant or product per unit time

Explanation:

6 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$