Answer:
8.55x10^22 molecules
Explanation:
From the question given, the following data were obtained:
Density = 1g/mL
Volume = 2.56mL
Mass =?
Density = Mass /volume
Mass = Density x volume
Mass = 1 x 2.56
Mass = 2.56g
Now let us convert this mass (i.e 2.56g) of water to mole
Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol
Mass of H2O = 2.56g
Number of mole of H2O=? Number of mole = Mass /Molar Mass
Number of mole of H2O = 2.56/18
Number of mole of H2O = 0.142mol
From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of H2O contains 6.02x10^23 molecules.
Now if 1mole of H2O contains 6.02x10^23 molecules, then 0.142mol of H2O will contain = 0.142 x 6.02x10^23 = 8.55x10^22 molecules
Answer : The volume of 6M NaOH stock solution is, 16.7 mL
Explanation :
To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:

where,
are the molarity and volume of NaOH stock solution.
are the molarity and volume of NaOH.
We are given:

Putting values in above equation, we get:

Thus, the volume of 6M NaOH stock solution is, 16.7 mL
Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is =

Now after adding extra 10 mL , the density becomes
.
Therefore, 
So the density decreases when we add more solution.
Explanation:
Moles of N2 = 35.0g / (28g/mol) = 1.25mol
Moles of H2 = 60.0g / (2g/mol) = 30.0mol
Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.
Moles of NH3 = 1.25mol * 2 = 2.50mol.
Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.
30.0mol - 1.25mol * 3 = 26.25mol.
Excess mass of H2
= 26.25mol * (2g/mol) = 52.5g.