The correct answer is approximately 1.95g. First, take (1.022*10^10)/(1.277*10^9). This is the same as time elapsed/half-life. This gives you 8, which is the number of half lives. Then multiply mass (500.3g) by (1/2)^8, resulting in 1.95g, which is the answer.
The standard addition equation is as followsI_(S+X) (V/V_O )=I_X+I_X/[X]_i [S]_4 (V_S/V_0 ) Here, [X]_i is the initial concentration of analyte, [S]_i is the initial concentration of standard, I_X is signal for analyte, I_(S+X) is signal for standard and analyte, V_0 is the initial volume, V_S is the added standard volume, and V is the total volume.Added volume of standard V_S is-23.3 mL. Initial volume of the sample V_0 is 10.00 mL. Initial concentration of standard ([S]_i) is 0.156 ng/mL.[X]_i= -[S]_i (V_S/V_0 )〖[X]〗_(i )= -(0.156 ng/mL)((-23.3 mL)/(10.00 mL))=0.363 ng/mL
Concentration of U(III) in ground sample is 0.363 ng/mL
Answer:
I' really don't know I'm sorry
V=4.8 L
c=5.0 mol/L
M(Mg)=24.3 g/mol
1) n(HCl)=cv
2) m(Mg)=M(Mg)n(HCl)/2
3) m(Mg)=M(Mg)cv/2
m(Mg)=24.3*5*4.8/2=291.6 g
You've answered your question it is 1 ton
