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raketka [301]
3 years ago
6

Each enzyme functions in a different area of the body, but both work to harness the power in carbohydrates and break these compl

ex molecules into simple sugars. Based on the information you have learned about the digestive system, what do you expect to be the optimal pH for the enzymatic action of pancreatic amylase?
Chemistry
2 answers:
mylen [45]3 years ago
7 0
The optimal pH is 6.9. The pancreatic amylase is a hydrolase secreted by the pancreas and will go into the alimentary canal. The working environment has a pH of 6.9. So the optimal pH for the enzyme needs to be 6.9.

Explanation:

The optimum pH scale for the catalyst activity of secretion enzyme ranges from half-dozen to seven. higher than and below this vary, the reaction rate reduces as enzymes get denaturated. The accelerator secretion enzyme is most active at pH scale 6.8.
Human duct gland alpha-amylase. II. Effects of hydrogen ion concentration, substrate and ions on the activity of the protein. refined human duct gland alpha-amylase (alpha-1,4-glucan 4-glucanohydrolase, EC 3.2.1.1) was found to be stable over a large vary of hydrogen ion concentration<span> values (5.0 to 10.5.</span>
Whitepunk [10]3 years ago
5 0
The optimal pH is 6.9. The pancreatic amylase is a hydrolase secreted by pancreas and will go into the alimentary canal. The working environment has a pH of 6.9. So the optimal pH for the enzyme needs to be 6.9.
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The equation for the photosynthesis reaction in which carbon dioxide and water react to form glucose is . The hear reaction is the difference between the bond dissociation energies in the products and the bond dissociation energies of the reactants

 

The reactant molecules have 12 C = O, 12 H - O bonds while the product molecules have 5 C - C, 7 C – O, 5 H – O, and 6 O = O bonds. The average bond dissociation energies for the bonds involved in the reaction are 191 for C = O, 112 for H – O, 83 C –C, 99 C – H, 86 C – O, 119 O = O.

 

Substitute the average bond dissociation energies in the equation for and calculate as follows

= [12 (C=O) + 12 (H-O)] – [5(C-C) + 7(C-H) + 7 (C-O) + 5(H-O) + 6(O=O)]

= [12x191 kcal/mol + 12x112 kcal//mol] – [5x83 kcal/mol + 7x99 kcal/mol + 7x86 kcal/mol + 5x112 kcal/mol + 6x119 kcal/mol]

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So, enthalpy change for the reaction is 652 kcal/mol or 2.73x10^3 kJ/mol

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How many atoms of magnesium metal are needed to produce 3.87 moles of magnesium oxide
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Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

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