Answer:
Rate of formation of SO₃
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒
-----------------------------(1)
Given that the rate of disappearance of oxygen =
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃
= ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒
= 2 x 3.64 x 10⁻³ M/s
⇒
= 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃
= 7.28 x 10⁻³ M/s
Heating up a reaction increases the speed of a reaction until the enzyme denatures.
<h3>What is enzyme denaturation?</h3>
Enzyme denaturation occurs when a biological protein catalyst does not work anymore due to a high temperature that alters its tridimensional conformation.
This cellular process (denaturation) is well known to be one of the main causes of enzymatic failure.
In conclusion, heating up a reaction increases the speed of a reaction until the enzyme denatures.
Learn more about enzymes here:
brainly.com/question/1596855
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The charge of a proton is positive.
Answer : The temperature of liquid is, 369.9 K
Explanation :
The Clausius- Clapeyron equation is :

where,
= vapor pressure of liquid at 373 K = 681 torr
= vapor pressure of liquid at normal boiling point = 760 torr
= temperature of liquid = ?
= normal boiling point of liquid = 373 K
= heat of vaporization = 40.7 kJ/mole = 40700 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:


Hence, the temperature of liquid is, 369.9 K
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.