The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.
(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.
Answer:
When a copper is drawn into wire the only change that occurs is change in its shape and size no change will take place into its composition that is the wires are still possessing the properties of copper metal. Thus, a physical change takes place when copper is drawn into wire.
<span>The correct answer is( A) blood.
when the buffer solution its PH value changes very little when a small amount
of strong acid or base is added to it, and here the bicarbonate buffering system is used to regular the PH of the blood that keeping the PH at nearly constant value by maintaining the original acidity or basicity of the solution.</span>
Answer:
C₅H₁₀O₅
Explanation:
1. Calculate the mass of each element in 2.78 mg of X.
(a) Mass of C

(b) Mass of H

(c) Mass of O
Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g
2. Calculate the moles of each element

3. Calculate the molar ratios
Divide all moles by the smallest number of moles.

4. Round the ratios to the nearest integer
C:H:O = 1:2:1
5. Write the empirical formula
The empirical formula is CH₂O.
6. Calculate the molecular formula.
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ

MF = (CH₂O)₅ = C₅H₁₀O₅
The molecular formula of X is C₅H₁₀O₅.
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.