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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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A sample of sugar (C12H22O11) contains
Lady_Fox [76]

4 moles of sugar.

Explanation:

A mole is defined as the amount of a substance contained in Avogadro's number of particles 6.02 x 10²³.

    1 mole of substance  = 6.02 x 10²³. molecules

Given that;

  the sample of sugar contains 1.505 x 10²³.molecules

  The number of moles in this amount of sugar is 4 moles

Learn more:

Number of moles brainly.com/question/13064292

#learnwithBrainly

5 0
3 years ago
if a girl holds a ball at waist level and then lifts the ball above her head and holds it there which of the following is she do
Zigmanuir [339]

Answer:

She is making kinetic energy when she lifts the ball and when the ball is above her head the ball gains potential energy.

4 0
3 years ago
Read 2 more answers
What is the result of a neutralization reaction between nitric acid (HNO3) and potassium hydroxide (KOH)?
scZoUnD [109]
HNO3+KOH = H2O+KNO3 . When nitric acid react with pottasuim hydroxide, the reaction will produce water (H20) and pottasuim trioxonitrate
3 0
3 years ago
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Consider these chemical equations. N2(g) + 3H2(g) → 2NH3(g) C(s) + 2H2(g) → CH4(g) 4H2(g) + 2C(s) + N2(g) → 2HCN(g) + 3H2(g) Whi
joja [24]
Multiply the second equation by 2
3 0
3 years ago
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A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
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