Answer:
I think it is A. the two forces are equal, or balanced, neither the table nor the book moves up or down.
Explanation:
Tome, the other options don’t make no sense
Answer:
The science project or the proposed test using the water as a control element in the experiment determines whether you choose to use deionized water or distilled water. Of the two, distilled water is the purest because the water undergoes boiling that kills off most organic contaminants.
Explanation:
HELP ME ON MINE PLZ
The options
Select one:
a. a 3- ion forms.
b. the noble gas configuration of argon is achieved.
c. the result is a configuration of 1s2 2s2 2p6.
d. the atom gains five electrons.
Answer:
c. the result is a configuration of 1s2 2s2 2p6.
Explanation:
Aluminium atom has atomic number of 13 , hence the number of electron is 13 for a neutral atom of aluminium. When aluminium atom reacts with other elements it usually gives out three electron to attain the octet configuration.
The cation representation of aluminium is Al3+ because it has loss three electron to attain the octet rule. Aluminium will be left with 10 electrons after losing 3 of it electrons. The electronic configuration will be represented as follows after losing three electrons;
1S² 2S² 2P∧6 .
At this stage the octet rule has been achieved as it will be represented as
2 8. The first energy shell now contains two electron and the second energy shell contains 8 electrons.
The configuration of Neon has been formed in the process.
Answer:
%Ionization = 1.63%
Explanation:
Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.
H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶
So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.
HONHNHOH => HONHNH⁺ + OH⁻
C(i) => 0.490M 0M 0M
ΔC => -x +x +x
C(eq) => 0.490 - x x x
≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100
Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]
1.3 x 10⁻⁶ = x²/0.490
=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴
=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%
Answer:
208.8018 g/Mol you conver the answer by weight and moles
