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givi [52]
2 years ago
11

Enzymes in the small intestine break down proteins, carbohydrates, and lipids. These enzymes are named after the substrate they

break down
A.Protease enzyme
B.Carbohydrate enzymes
C.Lipase enzymes

Chemistry
1 answer:
svetoff [14.1K]2 years ago
7 0

The enzymes and their respective substrates are as follows:

  • Protease enzymes such as trypsin and chymotrypsin break down proteins
  • Carbohydrate enzymes such amylase and maltase break down carbohydrates
  • Lipase enzyme breaks down lipids.

In the small intestine, a protease enzyme known as chymotrypsin breaks down protein, pancreatic amylase breaks down carbohydrates, while pancreatic lipase breaks down lipids.

More on biological enzymes can be found here: brainly.com/question/12194042

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Isotopes of the same element have different numbers of<br> protons<br> electrons<br> neutrons
MrRa [10]

Answer:

Protons and electrons

Explanation:

They have a different number of neutrons giving them a different atomic weight while having the same charge

6 0
3 years ago
A solution contains 11% by mass of sodium chloride. this means that __________.
d1i1m1o1n [39]

As a solution is contained with 11% by mass of sodium chloride, this will therefore conclude that the 100g of the solution is composed of the 11g of the sodium chloride. In which, the correct answer would be specified as letter c.

5 0
3 years ago
ANSWER QUICK
Novosadov [1.4K]

Answer:

1.8

Explanation:

8 0
3 years ago
Read 2 more answers
The number of atoms in 2 moles of aluminium is if the 1mole is 6.022*10 the power of 23 or 10^23
patriot [66]

Answer:

1.204 × 10²⁴ atoms

Explanation:

According to this question, one mole of aluminum (Al) atom contains 6.02 × 10²³ atoms.

If two moles of aluminum are given, this means that there will be 2 × 6.02 × 10²³ atoms of aluminum

2 × 6.02 × 10²3

= 12.04 × 10²³

= 1.204 × 10²⁴ atoms.

3 0
3 years ago
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
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