Answer: 462 g
Explanation:molar mass is M= 63.55 +2·(12.01+14.01)= 115.59 g/mol.
Mass m= n·M = 4.0 mol·115.59 g/mol= 462.36 g
Democritus was the first person to theorize the existence of atoms.
<span>Molecular formulas tell you how many atoms of each element are in a compound, and empirical formulastell you the simplest or most reduced ratio of elements in a compound. ... Also, many compounds with different molecular formula have the same<span>empirical formula</span></span>
The aim is to use less space while demonstrating the distribution of electrons in shells
If you want to depict how an atom's electrons are scattered across its subshells, an orbital notation is more suited.
This is due to the fact that some atoms have unique electronic configurations that are not readily apparent from textual configurations.
<h3>How does electron configuration work?</h3>
The placement of electrons in orbitals surrounding an atomic nucleus is known as electronic configuration, also known as electronic structure or electron configuration.
<h3>What sort of electron arrangement would that look like?</h3>
- For instance: You can see that oxygen contains 8 electrons on the periodic table.
- These 8 electrons would fill in the following order: 1s, 2s, and finally 2p, according to the aforementioned fill order. O 1s22s22p4 would be oxygen's electron configuration.
learn more about electronic configuration here
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Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.
Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g
From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>
