Answer:
0.0917 mol Co(CrO₄)₃
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
37.3 g Co(CrO₄)₃
<u>Step 2: Identify Conversions</u>
Molar Mass of Co - 58.93 g/mol
Molar Mass of Cr - 52.00 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol
<u>Step 3: Convert</u>
<u />
= 0.091662 mol Co(CrO₄)₃
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃
Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?




- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4

<span>plasmas that reach a temperature equal to their surroundings
----------------------=0uyjgyhjngytfgbhtfgbdc b gdf gb </span>
Answer:
Atmospheric nitrogen is not heavier than chemical nitrogen, largely because “chemical nitrogen” is ultimately derived from atmospheric nitrogen. On the other hand, you could be asking why the atomic mass of nitrogen is not the same as the mass of nitrogen gas; that's because gaseous nitrogen is diatomic, .
Explanation:
This is from Google.
Hope this helps :))
Answer:
47.8 g
Explanation:
Remember the equation for percent yield:
% yield = actual / theoretical
We're given two of the values in the question, so plug n' play:
0.945 = 45.2 / theoretical
theoretical = 47.8 g
Keep in mind you can use mass here without converting to moles because we're working with products only. If you were given a mass of reactants, you would need to convert to moles and using a balanced chemical equation find the corresponding moles of product produced.