1. c. 35
2. <span>d. 1.26 x 10^24 molecules
3. </span><span>d. 303.6 g</span>
Answer:
Phosphorus
Explanation:
Lewis Dot Structure is a structural representation of a molecule that shows valence electrons of a molecule.
The element in group 15, period 3 is Phosphorus (P). Phosphorus has an atomic number 15 and valence electrons are 5. So, the lewis structure will show 5 dots around P atom.
Answer:
![Ba\ percentage\ in\ Mass=4.8\%](https://tex.z-dn.net/?f=Ba%5C%20percentage%5C%20in%5C%20Mass%3D4.8%5C%25)
Explanation:
From the question we are told that:
Mass of mixture ![m=3.455g](https://tex.z-dn.net/?f=m%3D3.455g)
Mass of Barium ![m_b=0.2815g](https://tex.z-dn.net/?f=m_b%3D0.2815g)
Equation of Reaction is given as
![Ba2+ + H2SO4 => BaSO4 + 2 H+](https://tex.z-dn.net/?f=Ba2%2B%20%2B%20H2SO4%20%3D%3E%20BaSO4%20%2B%202%20H%2B)
Generally the equation for Moles of Barium is mathematically given by
Since
![Moles of Ba^{2+} = Moles of BaSO_4](https://tex.z-dn.net/?f=Moles%20of%20Ba%5E%7B2%2B%7D%20%3D%20Moles%20of%20BaSO_4)
Therefore
![Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol](https://tex.z-dn.net/?f=Moles%20of%20Ba%5E%7B2%2B%7D%20%3D%20%5Cfrac%7B0.2815%7D%7B233.39%7D%3D%200.0012061%20mol)
Generally the equation for Mass of Barium is mathematically given by
![Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g](https://tex.z-dn.net/?f=Mass%5C%20of%5C%20Ba%5E%7B2%2B%7D%20%3D%200.0012061%20%2A%20137.33%20%3D%200.1656%20g)
Therefore
![Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%](https://tex.z-dn.net/?f=Ba%5C%20percentage%5C%20in%5C%20Mass%3D%20%5Cfrac%7B0.1656%7D%7B%203.455%20%7D%2A%20100%25)
![Ba\ percentage\ in\ Mass=4.8\%](https://tex.z-dn.net/?f=Ba%5C%20percentage%5C%20in%5C%20Mass%3D4.8%5C%25)
<u>Answer:</u> The concentration of acetic acid in the vinegar is ![7.10\times 10^{-2}M](https://tex.z-dn.net/?f=7.10%5Ctimes%2010%5E%7B-2%7DM)
<u>Explanation:</u>
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
pH = 2.95
Putting values in above equation, we get:
![2.95=-\log[H^+]](https://tex.z-dn.net/?f=2.95%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=1.122\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.122%5Ctimes%2010%5E%7B-3%7DM)
The chemical equation for the ionization of acetic acid follows:
![CH_3COOH\rightleftharpoons H^++CH_3COO^-](https://tex.z-dn.net/?f=CH_3COOH%5Crightleftharpoons%20H%5E%2B%2BCH_3COO%5E-)
The expression of
for above equation follows:
![K_a=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
We are given:
![K_a=1.8\times 10^{-5}](https://tex.z-dn.net/?f=K_a%3D1.8%5Ctimes%2010%5E%7B-5%7D)
![[H^+]_{eq}=[CH_3COO^-]_{eq}=1.122\times 10^{-3}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7Beq%7D%3D%5BCH_3COO%5E-%5D_%7Beq%7D%3D1.122%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above expression, we get:
![1.8\times 10^{-5}=\frac{(1.122\times 10^{-3})\times (1.122\times 10^{-3})}{[CH_3COOH]}](https://tex.z-dn.net/?f=1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%281.122%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20%281.122%5Ctimes%2010%5E%7B-3%7D%29%7D%7B%5BCH_3COOH%5D%7D)
![[CH_3COOH]_{eq}=0.0699M](https://tex.z-dn.net/?f=%5BCH_3COOH%5D_%7Beq%7D%3D0.0699M)
So, the concentration of acetic acid = ![[CH_3COOH]_{eq}+[H^+]_{eq}=(0.0699+0.001122)=7.10\times 10^{-2}M](https://tex.z-dn.net/?f=%5BCH_3COOH%5D_%7Beq%7D%2B%5BH%5E%2B%5D_%7Beq%7D%3D%280.0699%2B0.001122%29%3D7.10%5Ctimes%2010%5E%7B-2%7DM)
Hence, the concentration of acetic acid in the vinegar is ![7.10\times 10^{-2}M](https://tex.z-dn.net/?f=7.10%5Ctimes%2010%5E%7B-2%7DM)
The correct option is (b)
NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').
Learn more about NANH2 here :-
brainly.com/question/12601787
#SPJ4