Answer:
D number of protons or atomic mass
 
        
                    
             
        
        
        
Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?
Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.
V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3
Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":
9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)
= 6.96 x 10^19 atoms of Ir-122 are present.
 
        
             
        
        
        
Answer: molecule of the kind normally found in living systems. Organic molecules are usually composed of carbon atoms in rings or long chains, to which are attached other atoms of such elements as hydrogen, oxygen, and nitrogen.
Explanation:
 
        
             
        
        
        
a) After adding 10 mL of HCl
first, we need to get moles of (CH3)3N = molarity * volume
                                                                 = 0.29 m * 0.025 L
                                                                 = 0.00725M moles
then, we need to get moles of HCl = molarity * volume
                                                          = 0.3625 m * 0.01L
                                                          = 0.003625 moles
so moles of (CH3)3N remaining    = moles of (CH3)3N - moles of HCl
                                                         = 0.00725 - 0.003625
                                                      = 0.003625 moles
and when the total volume = 0.01 L + 0.025L = 0.035 L
∴ [(CH3)3N] = moles remaining / total volume
                    = 0.003625 moles / 0.035L
                    = 0.104 M
when we have Pkb so we can get Kb :
pKb = - ㏒Kb
4.19 = -㏒Kb
∴Kb = 6.5 x 10^-5
when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]
and by using ICE table we assume we have:
[(CH3)3NH+] = X & [OH] = X 
and [(CH3)3N] = 0.104 -X
by substitution:
∴ 6.5 x 10^-5 = X^2 / (0.104-X)  by solving for X
∴X = 0.00257 M
∴[OH-] = X = 0.00257 M
∴POH = -㏒[OH]
           = -㏒0.00257
           = 2.5
∴ PH = 14 - POH
         = 14 - 2.5
         = 11.5
b) after adding 20ML of HCL:
moles of HCl = molarity * volume
                       = 0.3625 m * 0.02 L
                       = 0.00725 moles
  
the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:
and when the total volume = 0.02L + 0.025 = 0.045L
∴ [ (CH3)3NH+] = moles / total volume
                          = 0.003625 / 0.045L
                          = 0.08 M
 
when Ka = Kw / Kb 
and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14 
so, by substitution:
Ka = (1 x 10^-14) / (6.5 x 10^-5)
    = 1.5 x 10^-10
when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]
by substitution:
∴ 1.5 x 10^-10 = X^2 / (0.08 - X)  by solving for X
∴X = 3.5 x 10^-6  M
∴ [H+]= X = 3.5 x 10^-6 M
∴PH = -㏒[H+]
        = -㏒(3.5 x 10^-6)
       = 5.5
C) after adding 30ML of HCl:
moles of HCl = molarity * volume 
                       = 0.3625m * 0.03L
                       = 0.011 moles
and when moles of (CH3)3N neutralized = 0.003625 moles 
∴ moles of HCl remaining    = moles HCl - moles (CH3)3N neutralized
                                                = 0.011moles - 0.003625moles
                                                = 0.007375 moles
when total volume = 0.025L + 0.03L
                                = 0.055L
∴[H+] = moles / total volume
           = 0.007375 mol / 0.055L
            = 0.134 M
∴PH = -㏒[H+]
        = -㏒ 0.134
        = 0.87