Answer:
0.382g
Explanation:
Step 1: Write the reduction half-reaction
Al³⁺(aq) + 3 e⁻ ⇒ Al(s)
Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s
We will use the following relationships.
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons pass through the cell.
- The molar mass of Al is 26.98 g/mol.
The mass of Al produced is:

Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by
q=m C (T2-T1) ........1
Given : q= 2.1200 J
the initial temperature of gold, T1 = 22.0Celcius
the final temperature of gold, T2 = 1064.4Celcius
specific heat of gold = 0.131
putting values in eq 1:
⇒ 2.1200 = m × 0.131 × (1064.4-22)
⇒ 2.1200 = m × 0.131 × 1042.4
⇒ 2.1200 / 136.5544
⇒ 0.01552494829
Since 1g= 0.01552494829 Pounds
Mass of Gold = 267.165 × 0.01552494829
⇒ 4.1477228099
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Charge of nucleus is always positive whether it is element or isotope.
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M