Answer:
a. is an excel document. i uploaded as an attachment
b. concentration = 0.327
Explanation:
in the graph i uploaded for answer a, the x variable is conductivity, while the y variable is % salinity.
the calculated regression line was calculated using excel =
y = 0.057655X - 0.00891
for answer part b,
we take x = 5.82
when we put this into the y formular
y = 0.057655(5.82) - 0.00891
y = 0.3355521 - 0.00891
y = 0.3266421
this is ≈ 0.327
in conclusion, using the standard curve, the concentration of unknown salt is 0.327. the % salinity = 0.327
Answer:
Chemistry is the study of matter , its properties,how and why substance combine or separate to form other substance.
It's uses are
1) It is used in health care and beauty
2) Industries and transportation
3) Agriculture, science and technology
4) Cooking , cleaning, medicines etc
Answer:
The volume would be; 136.17 ml
Explanation:
Volume V1 = 150 mL
Temperature T1 = 20°C + 273 = 293 K
Pressure P1 = 758 - 17.54 = 740.46 torr
At STP;
Volume V2 = ?
Pressure P2 = 760 torr
Temperature T2 = 273 K
Using the general gas equation;
P1V1 / T1 = P2V2 / T2
Making V2 subject of formulae;
V2 = P1V1T2 / T1P2
Inserting the values we have;
V2 = 740.46 * 150 * 273 / 293 * 760
V2 = 136.17 ml
That’s picture right there should help u with your works
Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)
Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α
Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)
Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)
