Hey there!
2 moles will be produced.
In N₂ there are 2 nitrogen atoms. In NH₃ there is 1 nitrogen atom.
So, there will be twice as many moles produced because there will be twice as many molecules.
Hope this helps!
According to Dalton's Atomic Theory, the <em>Law of Definite Proportion is applied when a compound is always made up by a fixed fraction of its individual elements.</em> This is manifested by the balancing of the reaction.
The reaction for this problem is:
H₂ + Cl₂ → 2 HCl
1 mol of H₂ is needed for every 1 mole of Cl₂. Assuming these are ideal gases, the moles is equal to the volume. So, if equal volumes of the reactants are available, they will produce twice the given volumes of HCl.
Answer:
Biggest Radii V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺ Smallest Radii
General Formulas and Concepts:
- Periodic Trends: Atomic/Ionic Radii
- Coulomb's Law
Explanation:
The Periodic Trend for Atomic Radii is down and to the left. Therefore, the element with the largest radius would be in the bottom left corner of the Periodic Table.
Anions will always have a bigger radii than the parent radii. When we add e⁻ to the element, we are increasing the e⁻/e⁻ repulsions. This will cause e⁻ to repel themselves more and thus create more space, increasing the radii size.
Cations will always have smaller radii than the parent radii. When we remove e⁻ from the element, we are decreasing e⁻/e⁻ repulsions. Since there are less e⁻, there is no need for more space and thus decreases the radii size.
Since Cations are smaller than the parent radii, the more e⁻ we remove, the smaller it will become.
Therefore, the least removed e⁻ Vanadium would be the largest and the most removed e⁻ Vanadium would be the smallest.
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:
Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):
Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:
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