How many moles do you have in 37.3 g of Co(CrO4)3
1 answer:
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
You might be interested in
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01
A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite