<span>After many experiments and many different approaches to the question, the scientist may be able to develop a theory. The theory explains why nature behaves in the way described by the natural law. It answers not only the original question, but also any other questions that were raised during the process. The theory also predicts the results of further experiments, which is how it is checked. Theories are not the end of the process.</span>
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
Sucrose and other simple sugars may dissolve in water because they are polar molecules with an unequal charge distribution. Water is also quite polar, capable of forming weak, temporary connections with other polar compounds.
Salt dissolves into ions, with Na being positively charged and CL being negatively charged. Because water is highly polar (parts of the molecule are negatively charged while others are positively charged), the sodium ions are surrounded by water molecules, with the negatively charged component of the water molecules surrounding the NA ion. The Cl ion experiences the inverse effect.
<h3>
How does salt dissolve in water compared to sugar?</h3>
A solution's solute and solvent are two different types of substances that can dissolve one another. Different solvents have different levels of solubility for different solutes. For instance, sugar is far more soluble in water than salt. Even sugar, though, has a limit on how much may dissolve.
learn more about solubility refer:
brainly.com/question/23946616
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Answer:
1) The overlap of the p orbitals of the carbon-carbon π bond would be lost
Explanation:
Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.
In the structure of but-2-ene present in the attachment, we can see the two isomers, <em>cis</em> and<em> trans</em>. These isomers cannot be interconverted by rotation around the carbon-carbon double bond without breaking the π bond.
