Answer: 111g S
Explanation:
SO2 you have = 665 g
SO2 molar mass = 64.066 g/mol
S molar mass = 32.065 g/mol
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 %
Explanation:
here's the answer to your question
Answer:
0.15 L
Explanation:
You need to first find the volume of the container. You can do this by dividing the mass by the density. This will give you the mass in mL.
5.00 kg = 5,000 g
(5,000 g)/(1.00 g/mL) = 5,000 mL
5,000 mL = 5 L
Now, find the volume the seawater will take up.
(5,000 g)(1.03 g/mL) = 4854.4 mL
4854.4 mL = 4.85 L
Subtract the two volumes to find the volume that left unfilled.
5 L - 4.85 L = 0.15 L
Answer: 0.85075J/g.K
Explanation:
Mass of the material = 100g
Energy (Q) = 5104.5J
T1 = 20°C = 293K
T2 = 80°C = 353K
Formula for heat energy (Q) = mc ∇T
Q = mc∇T
∇T = T2 - T1
∇T = 353K - 293K = 60K
Q = mc∇T
C = Q / m∇T
C = (5104.5) / (100 * 60)
C = 0.85075 J/gK
The specific heat capacity of the material is 0.85075J/gK.
