This problem is easily solvable because radioactivity equations are common and well-established. The pseudo-first reaction is written below:
A = A₀(1/2)^(t/h)
where
A is the final amount
A₀ is the original amount
t is the time
h is the half life
5,000 = A₀(1/2)^(24,000/6,000)
Solving for A₀,
<em>A₀ = 80,000 atoms</em>
<h2>
Answer:</h2><h3>22226.026 g</h3><h2>
Explanation:</h2>
To get an approximate result, multiply the mass value by 454.
<em>hope</em><em> </em><em>this</em><em> </em><em>help</em><em>!</em>
Answer:
See the explanation below, please.
Explanation:
In the bunsen burner, the gas and air inlet can be regulated manually. In the case of metals (such as lithium and sodium in this case) they contain an electron in the latter in its external electronic configuration. They are characterized by transferring electrons easily and produce the emission of light when excited, producing flames of different colors in the lighter (orange for sodium and red / scarlet for lithium)
I think in drylands, drylands are particularly affected by desertification.
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
