Answer:
Mass = 40.4 g
Explanation:
Given data:
Mass in gram = ?
Volume of SO₂ = 14.2 L
Temperature = standard = 273 K
Pressure = standard = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 14.2 L = n × 0.0821 atm.L/ mol.K × 273 K
14.2 atm.L = n × 22.41 atm.L/ mol
n = 14.2 atm.L/22.41 atm.L/ mol
n = 0.63 mol
Mass of sulfur dioxide:
Mass = number of moles × molar mass
Mass = 0.63 mol × 64.1 g/mol
Mass = 40.4 g
Here we have to write a simple equation which describes the action of the enzyme catalase.
The equation is: The concentration of the complex [ES] = ![\frac{[E]0}{1+\frac{Km}{[S]} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5BE%5D0%7D%7B1%2B%5Cfrac%7BKm%7D%7B%5BS%5D%7D%20%7D)
Let us consider an enzyme catalyses reaction E + S ⇄ ES → E + P
Where E, S, ES and P are enzyme, substrate, complex and product respectively.
The concentration of the complex [ES] = , where 
is the Michaelis constant.
[E]₀ and [S] is the initial concentration of enzyme and concentration of substrate respectively.
HI is strong acid, so:
[H+] = [HI]
[H+] = 6 × 10^-3 M
pH = -log[H+]
pH = -log(6 × 10^-3) = 2,22
:-) ;-)
1.194 mol
(remember to use sig figs!)
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
