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Liula [17]
3 years ago
12

Which two scenarios illustrate the relationship between pressure and volume as described by Boyle’s law?

Chemistry
2 answers:
Airida [17]3 years ago
3 0

The correct answer is option 2 and 3.

The two scenarios that illustrate the relationship between pressure and volume as described by Boyle’s law are as follows:

2. The volume of an underwater bubble increases as it rises and the pressure decreases.

3. The pressure increases in an inflated plastic bag when the bag is stepped on.

According to Boyle's law, pressure of a gas is inversely proportional to its volume at constant temperature. This means that pressure rises as the volume increases and vice versa.

DaniilM [7]3 years ago
3 0

Answer:

2 and 3

Explanation:

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A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis o
MAVERICK [17]

Answer : The value of \Delta G_{rxn} is -49.6 kJ/mol

Explanation :

First we have to calculate the reaction quotient.

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

ATP(aq)+H_2O(l)\rightarrow ADP(aq)+HPO_4^{2-}(aq)

The expression for reaction quotient will be :

Q=\frac{[ADP][HPO_4^{2-}]}{[ATP]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Given:

[ATP] = 5.0 mM

[ADP] = 0.60 mM

[HPO_4^{2-}] = 5.0 mM

Now put all the given values in this expression, we get

Q=\frac{(0.60)\times (5.0)}{(5.0)}=0.60mM=0.60\times 10^{-3}M

Now we have to calculate the value of \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient = 0.60\times 10^{-3}

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-30.5kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (310K)\times \ln (0.60\times 10^{-3})

\Delta G_{rxn}=-49.6kJ/mol

Therefore, the value of \Delta G_{rxn} is -49.6 kJ/mol

6 0
3 years ago
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