This
reaction is called the electrolysis of water. The balanced reaction is:
2H2O = 2H2 + O2
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We are given the amount of water for the electrolysis reaction. This
will be the starting point of our calculation.
45.6 grams H2O (1 mol H2O / 18.02 g H2O) (1 mol O2 / 2 mol H2O) = 1.27 mol O2
V = nRT/P = </span><span>1.27 mol O2 (0.08206 atm L / mol K) (301 K) / 1.24 atm
V = 25.20 L O2</span>
D) the college students
Explanation:
In the experiment conducted by staffs of the nursing home, the group of college students that came for the experiment are the subjects of the experiment. A subject is a person that participates in a research.
- In this research, the college students are the human participants in the experiment.
- There must be a mutual agreement between the researcher and the subject because the subject is one that is being tested to see an outcome.
- A subject can quit a research if he/she deems it fit at any time.
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Answer:
(D) (CH3CH2)2NH
Explanation:
In order to decide which base is strongest we need to calculate its PKb
PKb = -log [Kb]
A large Kb value and small PKb value gives the strongest base
Compound Kb PKb
(A) C6H5NH2 - 4 x 10^-10 9.349
(B) NH3 1.76x 10^-5 4.754
(C) CH3NH2 4.4x 10^-4 3.357
(D) (CH3CH2)2NH 8.6x 10^-4 3.066
(E) C5H5N 1.7x10^-9 8.77
Clearly (CH3CH2)2NH is the strongest base.
Answer:
Yes
Explanation:
Masses for the three subatomic particles can be expressed in amu (atomic mass units) or grams. For simplicity, we will use the amu unit for the three subatomics. Both neutrons and protons are assigned as having masses of 1 amu each.
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9
