Answer: Copper(I) sulfate, also known as cuprous sulfate and dicopper sulfate, is the chemical compound with the chemical formula Cu2SO4 and a molar mass of 223.15 g mol−1. It is an unstable compound as copper(I) compounds are generally unstable and is more commonly found in the CuSO4 state. It is white in color at room temperature and is water-soluble. Due to the low-stability of the compound there are currently not many applications to date.
During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.
So there are some different indicator are used in table 2 as compared to the table 1.
- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2
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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.
Answer:
5.0 x 10⁹ years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of K-40 = 1.251 × 10⁹ years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
