Answer:
2 moles of Methane gas will occupy 44.8 L
Explanation:
If one mole of any gas occupies 22.4 L under certain conditions of temperature and pressure, and those conditions are assumed in this question, then we comfortably solve this problem as follows;
1 mole of Methane gas ---------------> 22.4 L
2 moles of Methane gas -------------->?
Cross and multiply, 2 moles of Methane gas = 2 X 22.4 L = 44.8L
Therefore, 2 moles of Methane gas will occupy 44.8 L, if the conditions of temperature and pressure are maintained.
OPTION A IS THE RIGHT SOLUTION.
Answer:
The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.
<u>Quick conversion chart of moles H2SO3 to grams</u>
1 moles H2SO3 to grams = 82.07908 grams
2 moles H2SO3 to grams = 164.15816 grams
3 moles H2SO3 to grams = 246.23724 grams
4 moles H2SO3 to grams = 328.31632 grams
5 moles H2SO3 to grams = 410.3954 grams
6 moles H2SO3 to grams = 492.47448 grams
7 moles H2SO3 to grams = 574.55356 grams
8 moles H2SO3 to grams = 656.63264 grams
9 moles H2SO3 to grams = 738.71172 grams
10 moles H2SO3 to grams = 820.7908 grams
The correct answer would be 2
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.
