Answer:
238,485 Joules
Explanation:
The amount of energy required is a summation of heat of fusion, capacity and vaporization.
Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)
m (mass of water) = 75 g
Lf (specific latent heat of fusion of water) = 336 J/g
C (specific heat capacity of water) = 4.2 J/g°C
∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C
Lv (specific latent heat of vaporization of water) = 2,260 J/g
Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J
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Answer:
7mL of sterile water is the initial amount of the concentrated solution is 3mL
Explanation:
In this problem, the vial must be <em>diluted </em>from 5mg/mL to 1.5mg/mL, that means the solution must be diluted:
5mg/mL / 1.5mg/mL = 3.33 times
If the initial amount of the drug in the vial is 3mL, the final volume must be:
10mL
That means the volume of water that should be added is:
10mL - 3mL:
<h3>7mL of sterile water is the initial amount of the concentrated solution is 3mL</h3>
PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-9)
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5
The concentration of OH- ions is 1.0*10^-5 M.