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Pachacha [2.7K]
3 years ago
12

If a 125 gms of radioactive element has a half life of 60 min how many half lives will it go through to become 3.90625 gms

Chemistry
1 answer:
Sonja [21]3 years ago
4 0

Answer:

  • 5

Explanation:

Let the number of half lives be x

<u>Solve this equation to find the value of x:</u>

  • 125*(1/2)ˣ = 3.90625
  • (0.5)ˣ = 3.90625 / 125
  • (0.5)ˣ = 0.03125
  • log (0.5)ˣ  = log 0.03125
  • x =  log 0.03125 / log 0.5
  • x = 5

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What orbital is represented by the transition metals in period four
andreev551 [17]

The name transition metal refers to the position in the periodic table of elements. The transition elements represent the successive addition of electrons to the d atomic orbitals of the atoms. In this way, the transition metals represent the transition between group 2 (2A) elements and group 13 (3A) elements.

8 0
3 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
Is each of these statements true? If not, explain why.(m) For most reactions, ΔH°rnx is lowered by a catalyst.
barxatty [35]

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.

learn more about catalyst refer

brainly.com/question/12507566

#SPJ4

3 0
1 year ago
How many electrons do all atoms want to have in their outer energy level?
34kurt

Answer: 8 electrons

Explanation:

4 0
3 years ago
All reversible reactions will reach chemical equilibrium
Fudgin [204]
True, all reversible reactions will reach chemical equilibrium. 
5 0
3 years ago
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