Answer:
Explanation:
Given , molarity of glycerol= 
Volume= 1 L.
Therefore, No of moles of glycerol= 
Now, volume of water needed, V=998.8 mL.
Density is given as= 0.9982 g/mL.
Therefore, mass of water = 
Now, molality=
Hence, this is the required solution.
<span>NaCH3COO (s) + HCl (aq) ---> HCH3COO (aq) + NaCl (s)</span>
Answer:
(A) The work done by the system is -101.325J
(B) The workdone by the system is -90.75J
Explanation:
(A) Workdone = -PΔV
Given that A = 100cm2 = 0.01m2
distance d = 10cm = 0.1m
ΔV= Area × distance
ΔV= 0.01 ×0.1
ΔV = 0.001m3
P= external pressure = 1atm = 101325Pa
Workdone = -0.001 × 101325
W= - 101.325Pa m3
1Pam3 = 1J
Therefore W = - 101.325J
The work done on the system is -101.325J
(B) Workdone = -PΔV
Given that A = 50cm2 = 0.005m2
distance d = 15cm = 0.15m
ΔV= Area × distance
ΔV= 0.005×0.15
ΔV = 0.00075m3
P=121kPa = 121000Pa
W= - 121000 × 0.00075
W= -90.75Pa m3
1Pam3 = 1J
W = - 90.75J
The woekdone by the system is -90.75J
It glow, so light energy go out of the system, exotermic
Answer:
A = 679.2955 ppm
Explanation:
In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:
A = A₀ e^-kt
This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:
k = ln2 / t₁/₂
Replacing the given data we have:
k = ln2 / 12.7
k = 0.0546
Now, let's get the concentration of Cu:
A = 845 e^(-0.0546*4)
A = 845 e^(-0.2183)
A = 845 * 0.8039
A = 679.2955 ppm
This would be the concentration after 4 hours
