Answer:
Average atomic mass of chlorine is 35.48 amu.
Explanation:
Given data:
Percent abundance of Cl-35 = 76%
Percent abundance of Cl-37 = 24%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (76×35)+(24×37) /100
Average atomic mass = 2660 + 888 / 100
Average atomic mass = 3548/ 100
Average atomic mass = 35.48 amu
Average atomic mass of chlorine is 35.48 amu.
N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹
n(Mg)=N/Nₐ
m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ
m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
Answer:
I believe 11 is B 12 is C 13 is B and 14 is C
Answer:
C
Explanation
His greatest contribution is the atomic model
