By 1.23 x 1024 you mean 10 to the power of 24 molecules? If so all you need to do is divide the number of molecules you have by Avagadros number, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance.
1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20
<h3>
Answer:</h3>
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂
[Given] 20 mol Al₂O₃
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Al₂O₃ → 4 mol Al
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4:Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
Since our final answer already has 1 sig fig, there is no need to round.
I think it could be C maybeee though
<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.
</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons; density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
