Answer:
The answer to the question is 7200
Answer is 2.19 g of radon will remain in the sample after 4 half-lives
When the experimental measurements diverge at very low values of the voltages. When the ratio of current to saturated current is very small. That is, i/I less than or equal to 1/100. When there is a finite current at Voltage
V = 0
Diffusion will have influence on current - voltage curve in the working electrode which is ionization detector.
The estimation under set of conditions that is possible, despite the spherical shape of the working electrode, to use linear diffusion equations to theoretically predict the current vs voltage curve expected in this experiment will be:
- When the experimental measurements diverge at very low values of the voltages
- when the ratio of current to saturated current is very small. That is, i/I less than or equal to 1/100
- When there is a finite current at Voltage V = 0
Where
i = electric current
I = saturated current
V = voltage supplied
The curve expected will therefore give exponential curve from positive to negative domain because of the diffusion of ions from different directions
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From the solubility rules, both reactions 1 and 2 lead to precipitates.
<h3>What is a precipitate?</h3>
The term precipitate refers to the solid that separates out of the reaction mixture . We know that the solubility of a substance in water is predicated on the solubility rules.
1) The reaction here is;
Fe(NO3)3(aq) + 3NaOH(aq) ------> Fe(OH)3(s) + 3NaNO3(aq) - The precipitate is Fe(OH)3 because only the hydroxides of group 1 elements are soluble in water.
2) The reaction is;
Pb(NO3)2(aq) + 2KI(aq) ----->PbI2(s) + 2NaNO3(aq) - The precipitate is PbI2 because most iodides are soluble except the iodides of Ag+, Hg+2, and Pb+2
Complete Ionic equation;
Pb^2+(aq) + 2NO3^-(aq) + 2K^+(aq) + 2I^-(aq) ------> PbI2(s) + 2NO3^-(aq) + 2K^+(aq)
Net ionic equation;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
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