I’d say A. They are theoretical in the fact that they have not been observed directly, but have been confirmed to exist through experimentation.
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
Answer:
0.37atm
Explanation:
Given parameters:
Initial pressure = 0.25atm
Initial temperature = 0°C = 273K
Final temperature = 125°C = 125 + 273 = 398K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use a derivative of the combined gas law;
= 
P and T are pressure and temperature
1 and 2 are initial and final values
= 
P2 = 0.37atm
T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
