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
Answer:
Exosphere
Explanation:
it is found at the end reaching outer space
I think it is C, because a covalent bond is a distribution of 2 atoms to 1 electron, meaning they are sharing and not exchanging, and the electronegravity would be above 1.7
Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)