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.
Explanation:
1)
Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M
A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution = 
Volume of the solution = 
Molarity of the NaOH solution after dilution = 
Volume of NaOH solution after dilution= 
A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.
The mass of the atom is equal to the sum of the number of protons and the number of neutrons. In a neutral atom, the number of protons is equal to the number of electrons. The atomic number meanwhile of an atom is equal to the number of protons of the atom.
what?? please reword this
Answer:
A sample of helium gas has a volume of 620mL at a temperature of 500 K. If we ... to 100 K while keeping the pressure constant, what will the new volume be?
Explanation:
