Given, half life of a certain radioactive element = 800 years.
Amount of substance remaining at time t = 12.5%
Lets consider the initial amount of the radioactive substance = 100%
Using the half life equation:
A = A₀(1/2)^t/t₁/₂
where A₀ is the amount of radioactive substance at time zero and A is the amount of radioactive substance at time t, and t₁/₂ is the half-life of the radioactive substance.
Plugging the given data into the half life equation we have,
12.5 = 100 . (1/2)^t/800
12.5/100 = (1/2)^t/800
0.125 = (0.5)^t/800
(0.5)^3 = (0.5)^t/800
3 = t/800
t = 2400 years
Thus the object is 2400 years old.
A catalyst reduces H°rnx in most reactions. The answer is false
<h3>Do catalysts reduce delta H?</h3>
By reducing the activation energy required for the reaction to occur, a catalyst just modifies the route used to go from reactants to products. However, because it doesn't alter the state of the products or reactants, delta H is unaffected.
A catalyst reduces a reaction's activation energy, enabling a chemical reaction to occur. The number of reactant particles with energy above the activation energy increases as the temperature of a reaction rises.
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The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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Answer:
Molar mass of solute: 300g/mol
Explanation:
<em>Vapor pressure of pure benzene: 0.930 atm</em>
<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>
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It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:
Moles in 78.11g of benzene are:
78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>
Now, mole fraction replacing in Raoult's law is:
0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.
As mole of solvent is 1:
0.9677× total moles = 1 mole benzene.
Total moles:
1.033 total moles. Moles of solute are:
1.033 moles - 1.000 moles = <em>0.0333 moles</em>.
As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:
10.0g / 0.033moles = <em>300g/mol</em>
Answer:
It emits 1.64 x 10⁻¹⁸J of energy
Explanation:
The n = 1 is a lower quantum level compared to n = 2.
When a hydrogen atom moves from a higher level to a lower one, it simply emits the energy difference between the two levels.
- If a hydrogen atom moves from a lower energy level to a higher one such as from 1 to 2, they absorb the energy difference to attain the new excited state.
- So, for an electron in the hydrogen atom to move from a higher energy level to a lower one, it must emit 1.64 x 10⁻¹⁸J of energy.
