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densk [106]
3 years ago
13

during a bread making process glucose is converted to ethanol and carbon dioxide causing the bread dough to rise by itself this

process takes longer than a human life span however when zymase an nzyme produced by yeast is added to bread the bread will ride in 15-20 min what is the function of zymase
Chemistry
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

Zymase is acting as a catalyst

Explanation:

Zymase is an enzyme that is naturally produced in yeast. Enzymes are biological catalysts that speed up the rates of reactions in living things.

Zymase catalyses the breakdown of glucose into ethanol and carbon dioxide, which causes the bread to rise.

Zymase speeds up this reaction, but is not physically changed itself. Therefore, it is a catalyst.

Actually, zymase represents a collection of enzymes in yeast, not just one!

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acetaminophen is a stronger acid than methanol, and sodium methoxide is a stronger base than. write an acid base equilibrium rea
torisob [31]

Answer:

The equilibrium shifts to the right that is the forward reaction.

Explanation:

The chemical compound known as "Acetaminophen" is a chemical compound that is generally known to a layman as Paracetamol and it belongs to the drug class known as anagelsics which helps in the treatment of pain or say in the reduction of pain. Acetaminophen has the chemical Formula to be C8H9NO2, with the Molar mass of 151.163 g/mol and Boiling point of 420 °C.

The reaction between Acetaminophen and sodium methoxide gives methanol and acetaminophen sodium salt. Therefore, the acid base equilibrium reaction of these species is given as;

C8H9NO2 + CH3ONa <========> CH3OH + acetaminophen sodium salt.

The equilibrium shifts to the right that is the forward reaction.

8 0
3 years ago
A fixed amount of gas in a rigid container is heated from 300 k to 900 k, which of the following responses best describes what w
Llana [10]

Answer: Option d. pressure increase by a factor of 3

Explanation:

P1 = P

T1 = 300k

T2 = 900K

P2 =?

Volume is constant.

P1/T1 = P2/T2

P/300 = P2/900

P2 x 300 = P x 900

P2 = (P x 900)/300

P2 = 3P

The pressure increased by a factor of 3

4 0
3 years ago
What is the balanced equation for tin dioxide + hydrogen ---&gt; tin +water
Nesterboy [21]
SnO2 + 2H2 >>> Sn + 2H2O
8 0
3 years ago
Why do noble gases have low freezing points?
Papessa [141]
This is because, only <span> weak van der Waals forces or weak London dispersion forces are present between the atoms of the </span><span>noble gases.

Hope this helps!</span>
6 0
3 years ago
Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&
Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ/mole

(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.5kJ/mole

(2) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol

(3) CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)     \Delta H_3=726.4kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0
3 years ago
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