The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Explanation:
According to the given data, we will calculate the following.
Half life of lipase 
= 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
= 
Initial fat concentration 
= 45 
= 45 mmol/L
Rate of hydrolysis 
= 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) = 
= 45 (1 - 0.80)
= 9
or, = 9 mmol/L
It is given that 
= 5mmol/L
Therefore, time taken will be calculated as follows.
t = ![-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7BK_%7Bd%7D%7Dln%5B1%20-%20%5Cfrac%7BK_%7Bd%7D%7D%7BV%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7BS_%7Bo%7D%7D%7BS%7D%29%20%2B%20%28S_%7Bo%7D%20-%20S%29%5D)
Now, putting the given values into the above formula as follows.
t =
= ![-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s }{K_{M} ln (\frac{45 mmol/L }{9 mmol/L }) + (45 mmol/L - 9 mmol/L )]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7Dln%5B1%20-%20%5Cfrac%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7D%7B0.07%20mmol%2FL%2Fs%0A%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7B45%20mmol%2FL%0A%7D%7B9%20mmol%2FL%0A%7D%29%20%2B%20%2845%20mmol%2FL%20-%209%20mmol%2FL%0A%29%5D)
= 
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.
To find this, we will use this formula:
Molar mass of element
------------------------------------ x 100
Molar mass of compound
So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.
Ca: 40.078 g/mol
N2 (there is two nitrogens): 28.014 g/mol
O6 (there are six nitrogens: 3 times 2): 95.994 g/mol
When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!
95.994
-----------
164.086
When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.
So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
Answer:
A
Explanation:
This is because in the graph shown line A has a quite greater impact of refraction than line B .
Hence, we can conclude that line A has the greater reaction at a faster rate.
The answer is A! Hoped it’s sure!
Explanation: A chemical reaction involves breaking bonds in the reactants, rearranging the atoms into new groupings (the products), and forming new bonds in the products.
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