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Nuclear reaction involves the nuclei of atoms. The nucleus can either split (as is the case with fission) of fuse as is the case with nuclear fusion. Remember the main particles in the nucleus are protons and neutrons.
Explanation:
In nuclear fission, atomic nuclei split into lighter atoms through loss of protons and neutrons (such as through loos of a beta particles - 2 protons and 2 neutrons). In the case of fusion, two atomic nuclei fuse under pressure (hence adding neutrons and protons) to create heavier atoms (2 hydrogens can fuse to form a helium). Electrons are not involved in nuclear reactions. Electrons are mainly involved in chemical reactions.
Learn More:
For more on nuclear fission and fusion check out;
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brainly.com/question/2005577
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Upper N upper H subscript 4 upper C l (s) right and left arrows stacked above each other upper N upper H subscript 3 (g) plus upper H upper C l (g)
Explanation:
The given equation is;
NH₄Cl ⇄ NH₃ + HCl
This equation is clearly different from the other ones.
- In the reactant going forward, there is a right and left arrows stacked above each other.
- The symbol is ⇄ and it is used to show reversibility of chemical reactions.
learn more:
Chemical reactions brainly.com/question/3953793
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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
The answer that correctly describes a reaction that forms a disaccharide from two monosaccharides is Galactose + glucose = lactose.
Cellulose is not a disaccharide, glycogen is not a monosaccharide, and sucrose isn't either.
