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Goryan [66]
2 years ago
14

Both the esophagus and the small intestine are involved in the digestion of food. The esophagus squeezes food into the stomach b

y wave-like muscle contractions. Peptidase enzymes in the small intestine break food molecules into smaller molecules.
Which statement best describes changes to food during digestion?

A. The muscle contractions result in physical changes, while the action of the peptidase results in chemical changes.

B. The muscle contractions result in chemical changes, while the peptidase results in physical changes.

C. The muscle contractions and the action of the peptidase both result only in physical changes.

D. The muscle contractions and the action in the peptidase both result only in chemical changes.
Chemistry
1 answer:
UkoKoshka [18]2 years ago
3 0

Answer:

The answer is: A

Explanation:

The wave-like contractions are to be considered a physical change as they do not take part in the breaking-down of the food. The peptidase is a chemical change because it is used to chemically break down the food to retrieve nutrients and such.

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Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres
goldenfox [79]

The answer is 6.1*10^-3 atm.

The pictures and explanations are there.

3 0
2 years ago
Determine the percent composition by mass of a 100g salt solution which contains 20g salt
prisoha [69]

Answer:

Mass of solution=100g

mass of salt=20g

so; mass of solute=80g

percentage composition =(mass of salt/total

mass) ×100

=  \frac{20}{100}  \times 100 \\  = 20\%

glad to help you

hope it helps

8 0
1 year ago
Read 2 more answers
Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution
Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}

or,

\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}

First we have to calculate the equivalent weight of solute.

Molar mass of solute PO_4^{3-} = 94.97 g/mole

\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq

Now we have to calculate the normality of solution.

\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L

Therefore, the normality of the solution is, 30.006 N

5 0
2 years ago
Select the correct set of quantum numbers (n, l, ml, ms) for the first electron removed in the formation of a cation for stronti
matrenka [14]

Answer:

5,0,0,-1/2

Explanation:

The quantum numbers are a way to characterize the electrons, and so, identify the region that it's more probable to find it (orbital). They are:

- Principal quantum number (n): represents the shell or level, and varies from 1 to 7, and are represented by the letter K, L, M, N, O, P, and Q.

- Azimuthal quantum number (l): represents the subshell or sublevel, and is represented by 0,1,2,3.., and for the letters s, p, d, f,...

- Magnetic quantum number (ml): represents the orbital. It varies from -l to +l passing by 0. Each orbital can have 2 electrons.

- Spin quantum number (ms): represents the spin of the electron. It can be +1/2 or -1/2.

The strontium has an atomic number equal to 38, by the Linus Pauling's diagram, the electronic distribution is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²

The valence electron is at the subshell 5s, which has only one magnetic quantum number: 0. Because it has 2 electrons, the first one has spin =1/2, and the other -1/2. So the first electron of the formation of cation has quantum numbers:

n = 5; l = 0; ml = 0; ms = -1/2

7 0
2 years ago
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
1 year ago
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