Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
Answer:
Atoms of each element contain a characteristic number of protons and electrons. The number of protons determines an element's atomic number and is used to distinguish one element from another.
Answer:
2KCl + F₂ → 2KF + Cl₂
Explanation:
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
2KCl + F₂ → 2KF + Cl₂
In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.
All other options are incorrect because mass is not conserved.
Mg₂ + LiBr ---> LiMg + Br
In this equation mass of magnesium is more on reactant side.
Na +O₂ ---> Na₂O
In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.
H₂O ---> H₂ + O₂
In this equation there is less oxygen on reactant side while more oxygen on product side.
Answer: The partial pressure of 
is 1.86 atm
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
The given balanced equilibrium reaction is,
Pressure at eqm. 0.973 atm 0.548atm x atm
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 1.86 atm
Thus, the partial pressure of is 1.86 atm
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