Answer:
Water contains two parts of hydrogen and one part oxygen. Therefore, during the electrolysis of water the amount of hydrogen has collected in one of the test tubes is double than that of the oxygen produced and collected in the other last tube.
Answer:
This question is incomplete
Explanation:
This question is incomplete, however, there are some basic things that be used to answer the completed question on your own. Flame test is a test that is used to identify metal ions in a compound. Although, not all metal ions produce a colour in a flame test.
In a flame test, a "clean wire loop" is dipped in an unknown solid/mixture of solids, the loop where the solids must have attached to is then placed in the tip of a blue flame (perhaps of a bunsen burner). A colour change/changes is then observed during the course of this process. Some popular metal ions and there colour in flame test are listed below
Lithium ion ⇒ red
Sodium ion ⇒ yellow
Potassium ion ⇒ lilac
Calcium ion ⇒ orange-red
Barium ion ⇒ pale-green
Copper ion ⇒ blue-green
rubidium ion ⇒ red-violet
Lead ion ⇒ gray white
The number of different colours observed will ultimately determine the number of elements exposed to the flame
Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
<u> E°/V</u>
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
<u> E°/V</u>
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41
</u>
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
</u>
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
Answer:
No
Explanation:
because Cu2+ has donated two electrons
Answer:
5 mol.
Explanation:
Equation of the reaction
2SO2 + 2H2O + O2 --> 2H2SO4
By stoichiometry, 2 moles of SO2 reacted with 2 moles of water and 1 mole of O2 to give 2 mole of sulphuric acid.
Number of moles:
5.0 mol SO2
4.0 mol O2
20.0 mol H2O
Calculating the limiting reagent,
5 mol of SO2 * 1 mol of O2/2 mol of SO2
= 2.5 mol of O2(4 mol of O2 is present)
5 mol of SO2 * 2 mol of H2O/2 mol of SO2
= 5 mol of H2O(20 mol of H2O)
SO2 is the limiting reagent.
Therefore, number of moles of H2SO4 = 5 mol of SO2 * 2 mol of H2SO4/2 mol of SO2
= 5 mol of H2SO4.