The answer is
Physical properties: Properties that do not change the chemical nature of matter
Chemical properties: Properties that do change tha chemical nature of matter
Examples of physical properties are: color, smell, freezing point, boiling point, melting point, infra-red spectrum, attraction (paramagnetic) or repulsion (diamagnetic) to magnets, opacity, viscosity and density. There are many more examples. Note that measuring each of these properties will not alter the basic nature of the substance.
Examples of chemical properties are: heat of combustion, reactivity with water, PH, and electromotive force.
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Answer:
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Explanation:
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hope this helps
The radius of the anion is 7.413 nm
<h3>How to calculate the force of attraction between charges</h3>
The force of attraction (F) is given by the formula:
- F = (1/4π∈r²)(Zc*e)(Za*e)
where:
∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m
Zc = charge on the cation = +2
Zc = charge on the anion = -2
e = charge on an electron = 1.602 * 10⁻¹⁹ C
r = interionic distance
r = rc + ra
where rc and ra are the radius of the cation and anion respectively
F = 1.64 * 10⁻⁸ N
Therefore based on the equation of force of attraction:
1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²
r² = 5.63 * 10⁻¹⁷
r = 7.50 nm
Since r = rc + ra
where rc = 0.087 nm
thus, ra = r - rc = 7.50 - 0.087
ra = 7.413 nm
Therefore, the radius of the anion is 7.413 nm
Learn more about ionic radius at: brainly.com/question/2279609
