132 grams x (1 mol / 44 grams) = 3 moles
<span>3 moles X (22.4 L/ 1 mol) = 67.2 </span><span>L</span>
Answer:The mole is important because it allows chemists to work with the subatomic world with macro world units and amounts. Atoms, molecules and formula units are very small and very difficult to work with usually. However, the mole allows a chemist to work with amounts large enough to use.
Explanation:
Answer:
The correct answer is - Statement I is true, Statement-II is false.
Explanation:
Answer:
Explanation:
8.61+5.779 = 14.389 = 1.4389 × 10^1
25 - 12.5 = 1.25 x 10^1
56.35 / 13.2 = 4.2689
I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>