Answer:
1.53x10^22 atoms of Au
Explanation:
To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).
<em>Moles Au:</em>
5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au
<em>Atoms of Au:</em>
0.0254 moles * (6.022x10²³ atoms / 1 mole) =
<h3>1.53x10^22 atoms of Au</h3>
From the calculations, the half life of the material is 6.5 days.
<h3>What is radioactivity?</h3>
The term radioactivity has to do with the spontaneous disintegration of a specie.
Uisng the formula;
N=Noe^-kt
N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq
No = amount initially present = 1.75 x 10^12 Bq
k = rate constant = ?
t = time taken = 55 days
Hence;
4.995 ×10^9 = 1.75 x 10^12e^-55k
4.995 ×10^9/1.75 x 10^12 = e^-55k
2.85 * 10^-3 = e^-55k
ln2.85 * 10^-3 = -55k
k = ln2.85 * 10^-3/-55
k = 0.1066 day-1
Half life = 0.693/ 0.1066 day-1
= 6.5 days
Learn more about radioactivity:brainly.com/question/1770619
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Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L
Answer:
2Cl——>Cl2+2e-
Explanation:
It shows an electron loss or gain
Answer:
The answer is 2.660 mol/l
Explanation:
Given: n= 0.0665, v= 25.00ml
Required: C
C (molarity)= n (of solute)/ v (of solvent) [ standard unit: mol/l]
First convert volume of solvent in its standard unit, i.e. litres(L)
v= 25.00ml/1000= 0.02500L
C = 0.0665 mol / 0.02500 L= 2.660 mL (In proper significant digits i.e. 4 sigdigs)
Therefore, The molarity of the sulfuric acid is 2.660 mol/L :)
