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.
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
Answer is: silicon isotope with mass number 28 has highest relative abundance, this isotope is the most common of these three isotopes.
Ar₁(Si) = 28; the average atomic mass of isotope ²⁸Si.
Ar₂(Si) =29; the average atomic mass of isotope ²⁹Si.
Ar₃(Si) =30; the average atomic mass of isotope ³⁰Si.
Silicon (Si) is composed of three stable isotopes, ₂₈Si (92.23%), ₂₉Si (4.67%) and ₃₀Si (3.10%).
ω₁(Si) = 92.23%; mass percentage of isotope ²⁸Si.
ω₂(Si) = 4.67%; mass percentage of isotope ²⁹Si.
ω₃(Si) = 3.10%; mass percentage of isotope ³⁰Si.
Ar(Si) = 28.086 amu; average atomic mass of silicon.
Ar(Si) = Ar₁(Si) · ω₁(B) + Ar₂(Si) · ω₂(Si) + Ar₃(Si) · ω₃(Si).
28,086 = 28 · 0.9223 + 29 · 0.0467 + 30 · 0.031.
Parantheses mean present of a subgroup or a group of polyatomic ions in a chemical reaction.
<u>Explanation:</u>
In a chemical reaction, sometimes, a compound may be composed of group of polyatomic ions with other ions. Some of the polyatomic ions are sulfate, carbonate, nitrate, hydroxide, bicarbonate, ammonia etc.
So these polyatomic ions are formed by combining two or more elements. Thus, if the number of polyatomic ions in a compound is more than one, then we use parathesis and write those polyatomic ions in it and write the number of polyatomic ions present in the compound as subscript of the parathesis.
For example, Fe₂(SO₄)₃
So here SO₄ is a polyatomic ion and in the compound 3 atoms of SO₄ is required to neutralize the compound and thus paranthesis is used. Thus, parantheses mean present of a subgroup or a group of polyatomic ions in a chemical reaction.
Answer:
Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces of Na₂SO₄.
