Mass of solute = 25.8 g
mass solution = 212 g
% =( mass of solute / mass solution ) x 100
% = ( 25.8 / 212 ) x 100
% = 0.122 x 100
= 12.2 %
Answer:
The nuclear fuel used in a nuclear reactor needs to have a higher concentration of the U 235 isotope than that which exists in natural uranium ore. U235 when concentrated (or "enriched") is fissionable in light-water reactors (the most common reactor design in the USA).
Explanation:
It will also be halved... but the relationship is with Kelvin not c. So 273 +273 = 546 K. half of that is 273 K, then subtract 273... you get 0 degrees c
Answer:
1.55 × 10²⁴ atoms Xe
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 57.5 L Xe at STP
[Solve] atoms Xe
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe