First, you have to convert mass to moles by making use of molecular weights [N = 14 g/mol and O = 16 g/mol)
A.
5.6/14 = 0.4 mol N
3.2/16 = 0.2 mol O
mol ratio = 2 mol N/1 mol O => N2O
B.
3.5/14 = 0.25 mol N
8.0/16 = 0.5 mol O
mol ratio = 1 mol N/2 mol O => NO2
C.
1.4/14 = 0.1 mol N
4.0/16 = 0.25 mol O
mol ratio = 2 mol N/5 mol O => N2O5
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2H20+1O2——> 2H2O should be the answer
Answer:
1.43 M
Explanation:
We'll begin by calculating the number of mole of the solid. This can be obtained as follow:
Mass of solid = 8.60 g
Molar mass of solid = 21.50 g/mol
Mole of solid =?
Mole = mass / molar mass
Mole of solid = 8.60 / 21.50
Mole of solid = 0.4 mole
Next, we shall convert 280 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
280 mL = 280 mL × 1 L / 1000 mL
280 mL = 0.28 L
Thus, 280 mL is equivalent to 0.28 L.
Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:
Mole of solid = 0.4 mole
Volume = 0.28 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.4 / 0.28
Molarity = 1.43 M
Thus, the molarity of the solution is 1.43 M.
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.
