Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:
Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:
I Hope this helps, greetings ... DexteR! =)
He worked with large numbers of plants.
4. Molar mass of silver m Ms~=108 g/mol
Hence there are n=54*(1/108)=0.5 mols of Silver in 54 grams of Silver.
5. 6.3*(108/1)=680.4g
6. Avogadro's number : Na~=6.022×10^23<span>. </span>
6.0*(6.022*10^23/1)=36.132*10^23 atoms
7. Molar mass of Krypton : Mk=84 g/mol
112/84=1.33 moles of Kr
8. 1.93*10^24*(1/(6.022×10^23))=3.2 moles KF
9. Molar mass of Silicon : Ms=28 g/mol
86.2*(1/28)*(6.022×10^23/1)=18.5*10^23 atoms of silicon
10. Molar mass of Magnesium : M1=24 g/mol
4.8*10^24*(1/(6.022×10^23))*(24/1)=191 g Mg
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
Answer:
S = 1.1 × 10⁻⁹ M
Explanation:
NaCl is a strong electrolyte that dissociates according to the following expression.
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.
We can find the molar solubility (S) of AgCl using an ICE chart.
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
I 0 0.15
C +S +S
E S 0.15+S
The solubility product (Ksp) is:
Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)
If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M
