The equation for carbon-14 emission by Radium-223 nuclei is given below:
<h3>What is radioactivity?</h3>
Radioactivity is the spontaneous decay of a substance with emission of radiation.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.
Learn more about radioactivity at: brainly.com/question/3603596
#SPJ1
C. 1.0 M Al2O3 would be the best answer
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
learn more about Ions:
brainly.com/question/13692734
#SPJ4
