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BartSMP [9]
3 years ago
5

Food molecules contain biochemical energy which is made available by

Chemistry
1 answer:
Juliette [100K]3 years ago
4 0

Food molecules contain biochemical energy which is made available by a process called respiration.

 Respiration is the process within cells by which living things break down food chemicals in their bodies and use them as a source of energy.

The proteins, lipids and polysaccharides that make up most of the food we eat must be broken down into smaller molecules before our cells can use them either as a source of energy or as building blocks for other molecules. This process is named catabolism and occurs in 3 stages.

Stage 1 is the enzymatic breakdown of food molecules in the digestion process  into their monomer subunits- amino acids, glucose and glycerol.

Stage 2 is the process of glycolysis where each molecule of glucose is converted to pyruvate.

Stage 3 is production of ATP,  the form of energy needed by the body to function. This stage takes place in the mitochondria of the cells. ATP is produced from conversion of pyruvate to acetylCoA in a process called the Citric Acid Cycle.



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An object with a mass of 3.2kg has a force of 16.3 newtons to the right and 6.7 newtons to the left applied to it. What is the r
Pepsi [2]

Answer:

3m/s²

Explanation:

Given parameters:

Mass of object  = 3.2kg

Force to the right = 16.3N

Force to the left  = 6.7N

Unknown:

Acceleration of the object  = ?

Solution:

To solve this problem, we use newtons second law of motion;

   Net force  = mass x acceleration

 Net force on object  = Force to the right  - Force to the left

 Net force  = 16.3N  - 6.7N  = 9.6N

    So;

         9.6  = 3.2 x a

              a = \frac{9.6}{3.2}   = 3m/s²

7 0
2 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

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

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
3. How many grams of potassium permanganate will react with 36 grams of iron(III)<br>hydroxide? ​
Mrac [35]

7 would be correct because I divide it

8 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST
bija089 [108]

Answer:

sorry i'm not a scientist but i think it's C

Explanation:

sorry again if i get it wrong for u :(

3 0
2 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
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