What determines the amount of chemical energy a substance has?

1 answer:

6 0

Chemical energy is the kind of energy stored in the bonds formed by atoms and molecules in chemical compounds and elements. This energy is released during a chemical reaction and heat is often given out in the process. These kind of reactions where heat is given out as a by product are called exothermic reactions.

The major factor that determines how much chemical energy a substance has is the mass of that substance. Mass is defined as the amount of matter in a substance.

The higher the mass of a substance, the more concentrated that substance is and subsequently the greater the number of atoms and molecules.

Logically, the higher the number of atoms and molecules then the greater the number of bonds in that substance and subsequently the more the amount of chemical energy stored therein.

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jok3333 [9.3K]

Using stoichiometry:

5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg

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The reaction is run at two temperatures where temperature 1 is lower than temperature 2. Which relationship is correct for eithe

Tasya [4]

Answer:

The answer is " $K_1 < K_2$ "

Explanation:

In the given question, the value of the $K=Ae^{-\frac{Ea}{RT}}$ and the $T_1$ is the rate value which is the constant that is $K_2>K_1$ . As per the temperature value when its increase rate is constantly increasing. $E_a$ is activation energy it is not dependent on the temperature that why the answer is $K_1 < K_2$ .

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2 years ago

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$