Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:
Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M
Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M
According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Answer: A: Explaining natural events using facts and data AND B:Developing theories using many lines of evidence
Explanation: trust me.
Molarity = moles of solute/volume of solution in liters.
The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.
(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.
Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:
(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
The word elliptical refers to an Oval.
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)