Molarity of concentrated nitric acid = 15.9 M
Volume of the stock solution to be prepared = 500.0 mL
Concentration of the stock that is to be prepared = 0.750 M
Calculating moles from molarity and volume of stock:
Calculating volume of concentrated nitric acid to be taken for the preparation of stock solution:
Converting L to mL:
= 23.6 mL
Volume of distilled water to be added to 23.6 mL of 15.9 M nitric acid to get the given concentration = 5000.0mL-23.6mL=976.4 mL
Therefore, 976.4 mL distilled water is to be added to 23.6 mL of 15.9 M nitric acid solution to prepare 500.0 mL of 0.750M nitric acid.
Answer:
3.9%
Explanation:
Exact melting point = 123°C
Measured melting point = 128°C
% error; |measured - exact|/exact × 100
% error = |123 -128|/128 × 100
%error = 3.9%
Percentage error in temperature measurement = 3.9 %
The moles of potassium that you would need to prepare 1200 g of 5% potassium sulfate solution is 1.538 moles
calculation
calculate the mass potassium using the below formula
%M/M = mass of the solute(potassium)/mass of the solvent (potassium sulfate solution)
let the mass of potassium be represented by Y
then convert % into fraction = 5/100
5/100 = Y/1200
cross multiplication
100y = 6000
divide both side by 100
Y= 60 g
moles of potassium =mass/molar mass
= 60/39=1.538
Answer:
The answer to your question is: 70.7 %
Explanation:
Equation
Xe + 2F₂ ⇒ XeF₄
limiting reactant = Xe
Xe is the limiting reactant because the ratio is:
theoretical = 131/ 76 = 1.72 g
experimental ratio = 130/100 = 1.3 the amount of F increased.
131.3 g of Xe ------------------ 207 g of XeF₄
130 g of Xe ------------------- x
x = (130 x 207) / 131.3
x = 205 g of XeF₄
% yield = 145 / 205 x 100
% yield = 70.7
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
