Acetaminophen as a chemical formula of C8H9NO2. The molar
masses are:
C8H9NO2 = 151.163 g/mol
C = 12 g/mol
H = 1 g/mol
N = 14 g/mol
O = 16 g/mol
<span>TO get the mass percent, simply multiply the molar mass of
each elements with the number of the
element divide by the molar mass of acetaminophen, that is:</span>
%C = [(12 * 8) / 151.163] * 100% = 63.50%
%H = [(1 * 9) / 151.163] * 100% = 5.954%
%N = [(14 * 1) / 151.163] * 100% = 9.262%
<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>
Answer:
Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution
Explanation:
There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:
1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%
2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%
When clearing X and Y :
(X mL 6% + Y mL 1% = 100 mL 5%) (-15,7)
X mL 94% + Y mL 99% = 100 mL 95%
_______________________________
- Y 0.83 = 16.5
Y = 19.9 mL 1% solution
Replace Y in anyone equation and X = 80 mL 6% solution
I hope to see been helpful
Answer:
0.290 atm is the pressure of the system after 7.7min
Explanation:
The general first-order rate constant is:
ln [A] = -kt + ln [A]₀
<em>Where [A] is concentration of A in time t,</em>
<em>K is rate constant, 3.2x10⁻⁴s⁻¹</em>
<em>[A]₀ is initial concentration = 0.336atm.</em>
<em />
7.7 min are:
7.7min * (60s / 1min) = 462s
Solving:
ln [A] = -kt + ln [A]₀
ln [A] = -<em>3.2x10⁻⁴s⁻¹*462s</em> + ln [0.336atm]
ln [A] = -1.238
[A] =
<h3>0.290 atm is the pressure of the system after 7.7min</h3>
<em />
Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= 
=14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5= 
14.5 x (0.15 M) x 
= (CH₃OH)
0.2818 M = (CH₃OH)
Answer:
83°C
Explanation:
The following were obtained from the question:
M = 40g
C = 4.2J/g°C
T1 = 91°C
T2 =?
Q = 1300J
Q = MCΔT
ΔT = Q/CM
ΔT = 1300/(4.2x40)
ΔT = 8°C
But ΔT = T1 — T2 (since the reaction involves cooling)
ΔT = T1 — T2
8 = 91 — T2
Collect like terms
8 — 91 = —T2
— 83 = —T2
Multiply through by —1
T2 = 83°C
The final temperature is 83°C
