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3 years ago

How many atoms are there in 1.3 x 10^22 molecules of N(O)2

2 answers:

5 0

1 molecule of NO2 included N atom and 2 oxygen atoms, so a total of 3 atoms per molecule.

1.3 x 10^22 times 3 = 3.9 x 10^22 atoms

nikitadnepr [17]3 years ago

4 0

Answer:

3.9×10^22

Explanation:

I hope it is right

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If a reaction is reversible what are the relative amounts of reactant and product

Alina [70]

If a reaction is reversible, then it will attain the phase of Equilibrium and at that phase, the Amount of Reactants and Products would be: Equal

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2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

pogonyaev

Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

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