The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %
Moles of O₂ = 3.92
<h3>Further explanation</h3>
Given
V=1 L
T=-118 °C + 273 = 155 K
P = 49.77 atm
Required
moles of O₂
Solution
the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
Input the value :
n= PV/RT
Answer:
The flocculation basin often has a number of compartments with decreasing mixing speeds as the water advances through the basin. ... This compartmentalized chamber allows increasingly larger floes to form without being broken apart by the mixing blades.
Phenolphthalein is not a good indicator to use for a titration for a solution that has a ph of 6.0 at the equivalence point because the color change of the solution at this pH level is not sharp. It changes the color of the solution to pink starting from pH 8.3 to 10. A pH level lower than 8.3 would only show a colorless solution. Thus, you would not be able to distinguish whether the solution has reached its equivalence point at pH 6.0. It is best to use this indicator for a system that is using a strong base titrated with a weak acid.
Answer:
The sample will be heated to 808.5 Kelvin
Explanation:
Step 1: Data given
Volume before heating = 2.00L
Temperature before heating = 35.0°C = 308 K
Volume after heating = 5.25 L
Pressure is constant
Step 2: Calculate temperature
V1 / T1 = V2 /T2
⇒ V1 = the initial volume = 2.00 L
⇒ T1 = the initial temperature = 308 K
⇒ V2 = the final volume = 5.25 L
⇒ T2 = The final temperature = TO BE DETERMINED
2.00L / 308.0 = 5.25L / T2
T2 = 5.25/(2.00/308.0)
T2 = 808.5 K
The sample will be heated to 808.5 Kelvin
