Glucose and Galactose both have the same molecular formula, C6H12O6, but in the body, galactose must be first converted to glucose to make energy. The difference<span> is their </span>structures
Answer:
2
Explanation:
In two reactions energy is released.
1) C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂ + heat
It is cellular respiration reaction.It involves the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.
Glucose + oxygen → carbon dioxide + water + 38ATP
2) 2H₂ + O₂ → 2H₂O ΔH = -486 kj/mol
The given reaction is formation of water. In this reaction oxygen and hydrogen react to form water and 486 kj/mol is also released.
The reaction in which heat is released is called exothermic reaction.
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Answer:
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<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Among the given choices, the right statement is that computers are advantageous because it holds a larger amount of data over graphing calculators. Calculators are easier to transport and has lower power consumption. Answer here is D.
