Answer:
Option D is correct = 8.12 grams of NaCl
Explanation:
Given data:
Moles of sodium chloride = 0.14 mol
Mass of sodium chloride = ?
Solution:
Formula:
Number of moles = mass of NaCl / Molar mass of NaCl
Molar mass of NaCl = 58 g/mol
Now we will put the values in formula.
0.14 mol = Mass of NaCl / 58 g/mol
Mass of NaCl = 0.14 mol × 58 g/mol
Mass of NaCl = 8.12 g of NaCl
Thus, 0.14 moles of NaCl contain 8.12 g of NaCl.
Mass of sodium thiosulfate 
is 110. g
Volume of the solution is 350. mL
Calculating the moles of sodium thiosulfate:
= 0.696 mol
Converting the volume of solution to L:
Finding out the concentration of solution in molarity:
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Answer:
Name, Thallium. Symbol, Tl. Atomic Number, 81. Atomic Mass, 204.3833 atomic mass units. Number of Protons, 81. Number of Neutrons, 123. Number of Electrons, 81 ... It has a metallic luster when it is first exposed to air but it tarnish quickly
Explanation:
Hope this helped :)
