Answer:
35.453 good luck with your work
it is a great change because the salt goes in and turn the water to salt water
Answer:
A. 96.3 mg/dL
Absolute error: 5.7 mg/dL
Relative error: 5.6%
B. 97.2 mg/dL
Absolute error: 4.8 mg/dL
Relative error: 4.7%
C. 104.8 mg/dL
Absolute error: 2.8 mg/dL
Relative error: 2.7%
D. 111.5 mg/dL
Absolute error: 9.5 mg/dL
Relative error: 9.3%
E. 110.5 mg/dL
Absolute error: 8.5 mg/dL
Relative error: 8.3%
Explanation:
The formula for the absolute error is:
Absolute error = |Actual Value - Measured Value|
The formula for the relative error is:
Relative error = |Absolute error/Actual value|
In your exercise, we have that
Actual Value = 102.0 mg/dL
A. 96.3 mg/dL:


B. 97.2 mg/dL


C. 104.8 mg/dL


D. 111.5 mg/dL


E. 110.5 mg/dL


Answer:
The value of
of the an ethylamine is
.
Explanation:
The pH of the solution = 12.067
The pOH of the solution = 14 - pH =14-12.607 =1.933
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.933=-\log[OH^-]](https://tex.z-dn.net/?f=1.933%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.0117 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.0117%20M)

Initially
0.342 M 0 0
At equilibrium
(0.342-x) x x
The value of x = ![[OH^-]=0.0117 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.0117%20M)
The expression of
is given as:
![K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BC_2H_5NH_3%5E%7B%2B%7D%5D%5BOH%5E-%5D%7D%7B%5BC_2H_5NH_2%5D%7D)


The value of
of the an ethylamine is
.
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