From the concept of half- life, it would take 121.88 days for a 28.0 g sample of Iron-59 to decay to 7.00 g.
The process of determining how long it will take for an element to decay to half of its initial quantity is known as half-life. The half-life of Iron-59 is 44 days.
The half-life formula is given as: A = A₀(1/2)^(t/t₁/₂) Where,
A₀ is the initial amount.
A is the amount after some time t
T₁/₂ is the half-life of the element.
t is the time taken
Using the above formula, we can solve for t.
Initially, the mass of the Iron-59 sample is A₀ = 28.0 g, and its final mass is A = 7.00 g.
So, the initial amount of Iron-59 is A₀ = 28.0 g.
Using the half-life formula, we get:
A = A₀(1/2) ^(t/t₁/₂)
Putting the given values:
A/A₀ = (1/2) ^(t/T₁/₂)
7.00/28.0 = (1/2) ^(t/44)
1/4 = (1/2) ^(t/44)
Take the natural log of both sides of the equation
ln (1/4) = ln [(1/2) ^(t/44)]
ln (1/4) = (t/44) ln (1/2)
Solve for t
ln t = (ln (1/4)) / (ln (1/2))
= 2.77 × 44
= 121.88 days
So, it would take 121.88 days for a 28.0 g sample of Iron-59 to decay to 7.00 g.
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name a substance which can oxidize i- to i2, but cannot oxidize br- to br2
The substance that can oxidize I-to-I2 but cannot oxidize Br-to-Br2 is chlorine. Chlorine can be used as an oxidizing agent to convert I- to I2, but it is not capable of oxidizing Br- to Br2.
This is due to the relative strengths of the halogens. Chlorine is a stronger oxidizing agent than iodine, but bromine is stronger than both chlorine and iodine. Therefore, chlorine is capable of oxidizing iodide ions to iodine, but it cannot oxidize bromide ions to bromine because bromine is a stronger oxidizing agent than chlorine.
In the presence of iodide ions (I-), chlorine (Cl2) can oxidize iodide ions to produce iodine (I2) and chloride ions (Cl-). 2 I- (aq) + Cl2 (aq) → 2 Cl- (aq) + I2 (s)In the presence of bromide ions (Br-), chlorine (Cl2) is unable to oxidize bromide ions to produce bromine (Br2) and chloride ions (Cl-). 2 Br- (aq) + Cl2 (aq) → no reaction
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which assumptions can be applied for the isothermal processes of o2 (l, 1 atm) → o2 (l, 1000 atm)?
The ideal gas law equation can be used to make certain assumptions about the isothermal processes of O2 (l, 1 atm) to O2 (l, 1000 atm).The assumptions for the isothermal processes of O2 (l, 1 atm) to O2 (l, 1000 atm) are as follows:
1. The temperature remains constant since the process is isothermal.2. The system is closed and therefore the number of O2 molecules remains the same.3. There is no change in the internal energy of the system since the process is isothermal.4. The gas is assumed to be ideal which means that it follows the ideal gas law equation.5. There is no change in the volume of the system since the process is isothermal and the system is in a liquid state.
The ideal gas law equation can be expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At constant temperature, the ideal gas law equation can be simplified to PV = constant.Using the ideal gas law equation, the initial pressure can be calculated as P1 = (nRT)/V1 and the final pressure can be calculated as P2 = (nRT)/V2.
Since the temperature remains constant, the equation can be simplified to P1V1 = P2V2.The above assumptions and equation are applicable for the isothermal processes of O2 (l, 1 atm) to O2 (l, 1000 atm). The ideal gas law equation can be used to calculate the pressures and volumes at different stages of the isothermal process.
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