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
Noble gases.
group 18. elements that are all unreactive .
fixed naming!
The volume of the gas that occupy at STP is 165. 28 cm^3
calculation
by use of combined gas law that is P1V1/T1=P2V2/T2, where
P1=84.6 kpa
T1=23.5 +273=296.5 K
V1=215 cm^3
At STP T= 273 K and P= 101.325 Kpa
therefore p2 = 101.325 Kpa and T2 = 272 K V2=?
by making V2 the subject of the formula V2 =T2P1V1/P2T1
V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3
C
I have had this question on a test before!! Hope this helps
The answer is carbon dioxide