Answer: C. Electrochemical cells involve oxidation-reduction reactions.
Explanation: Oxidation occurs at the anode, and reduction occurs at the cathode.
Answer is: identity of the metal is gold (Au).
ω(Cl) = 35.06% ÷ 100%.
ω(Cl) = 0.3506; mass percentage of chlorine.
If we take 100 grams of the compound:
m(Cl) = ω(Cl) · m(compound).
ω(Cl) = 0.3506 · 100 g.
ω(Cl) = 35.06 g.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 35.06 g ÷ 35.45 g/mol.
n(Cl) = 0.99 mol; amount of substance.
In molecule MCl₃: n(M) : n(Cl) = 1 : 3.
n(M) = 0.33 mol; amount of unknown metal.
M(M) = m(M) ÷ n(M).
M(M) = (100 g - 35.06 g) ÷ 0.33 mol.
M(M) = 196.8 g/mol; molar mass of the gold.
Can you translate to english?
Answer:
1.14 atm and 1.139 mol
Explanation:
The <em>total pressure</em> of the container is equal to the <u>sum of the partial pressure of the three gasses</u>:
- P = Poxygen + Pnitrogen + Pcarbon dioxide
- 2.50 atm = 0.52 + 0.84 + Pcarbon dioxide
Now we <u>solve for the pressure of carbon dioxide</u>:
- Pcarbon dioxide = 1.14 atm
To c<u>alculate the number of CO₂ moles </u>we use <em>PV=nRT</em>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 32 °C ⇒ 32 + 273.16 = 305.16 K
1.14 atm * 25.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305.16 K
