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amid [387]
3 years ago
11

Just as heat can denature enzymes, so can a change in the pH. Enzymes that normally work in an acidic environment in the body ca

n become denatured if the pH becomes too alkaline. And enzymes that normally work in alkaline environments in the body can become denatured if the pH becomes too acidic. Taking an antacid to reduce the acidity (and increase the pH) of the stomach and to relieve heartburn can make a person more comfortable. What would be the major drawback to taking too many antacids?
Chemistry
2 answers:
erastovalidia [21]3 years ago
6 0

Answer:

Too much antacid which then cause a major decrease in acidity which would cause the enzymes in the stomach (which operate optimally at high acidity) to be denatured.

Explanation:

Enzymes in the stomach function optimally at high acidity. Antacids are generally basic solution which reduces the acidity in the stomach hence relieve heartburn. Excess antacid will cause the operating conditions to be basic, hence lead to denature of the enzymes.

nikitadnepr [17]3 years ago
5 0

Answer:

It will create an alkaline environment and can lead to denaturation of acidic enzymes.

Explanation:

<em>Taking too much antacids will significantly increase the pH of the stomach and create an alkaline environment within it. Consequently, enzymes that normally work in acidic environments in the body can become denatured or rendered inactive. This will create another problem for the individual.</em>

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Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

N_2+2O_2\rightarrow 2NO_2

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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