Answer:

Explanation:
<em>Ferrous Sulphate</em>
<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>
<em>, Sulphur Dioxide </em>
<em> as well as Ferric Oxide </em>
<em>.</em>
<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>
<em>Hence,</em>

<em>Now,</em>
<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>
<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>
.
<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>
<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>
<em> </em>
<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>
Answer:
The pOH = 1.83
Explanation:
Step 1: Data given
volume of the sample = 25.0 mL
Molarity of hydrocyanic acid = 0.150 M
Molarity of NaOH = 0.150 M
Ka of hydrocyanic acid = 4.9 * 10^-10
Step 2: The balanced equation
HCN + NaOH → NaCN + H2O
Step 3: Calculate the number of moles hydrocyanic acid (HCN)
Moles HCN = molarity * volume
Moles HCN = 0.150 M * 0.0250 L
Moles HCN = 0.00375 moles
Step 3: Calculate moles NaOH
Moles NaOH = 0.150 M * 0.0305 L
Moles NaOH = 0.004575 moles
Step 4: Calculate the limiting reactant
0.00375 moles HCN will react with 0.004575 moles NaOH
HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH
Step 5: Calculate molarity of NaOH
Molarity NaOH = moles NaOH / volume
Molarity NaOH = 0.000825 moles / 0.0555 L
Molarity NaOH = 0.0149 M
Step 6: Calculate pOH
pOH = -log [OH-]
pOH = -log (0.0149)
pOH = 1.83
The pOH = 1.83
Answer:
The required volume of hexane is 0.66245 Liters.
Explanation:
Volume of octane = v=1.0 L=
Density of octane= d = 
Mass of octane ,m= 
Moles of octane =
Mole percentage of Hexane = 45%
Mole percentage of octane = 100% - 45% = 55%

Total moles = 11.212 mol
Moles of hexane :

Moles of hexane = 5.0454 mol
Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g
Density of the hexane,D = 
Volume of hexane = V

(1 cm^3= 0.001 L)
The required volume of hexane is 0.66245 Liters.
Solution:
After the reaction of mixture is worked-up Washing three times the organic with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.
And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.
Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14