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.
Answer:
The International Bottled Water Association (IBWA), Alexandria, Va., and Beverage Marketing Corporation (BMC), New York, recently released 2015 bottled water statistics showing that Americans' consumption of bottled water increased by 7.9 percent and bottled water sales were up 8.9 percent from the previous year.
Explanation:
The volume of hydrogen gas that evolved is calculated as follows
by use of ideal gas equation
that is PV = nRT
P=745 mm hg
V= ?
R(gas constant)= 62.36 L.mm hg/mol.k
T= 20 + 273 = 293 k
n=number of moles which is calculated as follows
find the moles of Na used
= 0.52/23=0.023 moles
write the reacting equation
2Na +2H2O =2NaOH +H2
by use of reacting ratio between Na : H2 which is 2:1 therefore the mole of H2 = 0.023/2 =0.0115 moles
by making the volume the subject of the formula
v=nRT/P
V= (0.0115 x 62.36 x 293) / 745 = 0.283 L
Answer:
B) 1.65
Explanation:
See the image uploaded for the percent error formula
Work:
Theoretical Value - Accepted Value/Accepted Value * 100
= (24.7-24.3)/24.3 * 100
= 0.4/24.3 * 100
= 0.0164609053 * 100
= 1.64609053
Can be rounded to 1.65%
Amount of silver nitrate taken = 269.μmol 
Volume of the solution = 300. mL
Concentration of a solution is generally expressed in terms of molarity. Molarity is defined as the moles of a substance present per liter of the solution.
We want the concentration in millimoles/L.
Converting μmol to millimol solute:
μ
= 0.269 millimol
Volume from mL to L: 
Therefore concentration of the chemist's solution = 
