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8_murik_8 [283]

3 years ago

9

The titration of an acid (H2A) with LiOH solution generates the following titration curve. What are the main components (more th

an half of the initial amount of H2A, besides H2O) at equivalence point 1 (EP1) and equivalence point 2 (EP2)?
Main Components at EP1 Main Components at EP2
A H2A and OH− A2− and OH−
B HA− and Li+ A2− and Li+
C HA− and OH− HA− and Li+
D H2A and Li+ HA− and OH−

1 answer:

faust18 [17]3 years ago

3 0

Answer:

B) HA⁻ and Li⁺ A²⁻ and Li⁺

Explanation:

Titration of acid H₂A with LiOH solution.

At first equivalent point

H₂A + LiOH → HA⁻ + Li⁺ + H₂O

Main component at (EP1) => HA⁻ and Li⁺

At second equivalence point

HA⁻ + LiOH → A²⁻ + Li⁺ + H₂O

Main component at (EP2) => A²⁻ and Li⁺

Therefore, the correct answer is B

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An ion of iron has an 26 protons, 30 neutrons, and 23 electrons. What are its atomic number, atomic mass, and net charge?

sleet_krkn [62]

Answer:

Atomic number 26,atomic mass 26+30=56 dalton and net charge is 3+

Explanation:

The total number of proton present in an atom is known as the atomic number of that atom.From that point of view the atomic number of iron ion is 26.

The total number proton and neutrons present in the nucleus of an atom or ion is termed as atomic mass.From that point of view the the atomic mass of iron ion is 26+30=56 dalton

According to the given question iron ion contain 3 more protons than electrons as a result the iron ion will contain 3 unit positive charge.

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3 years ago

Vitamin C contains only carbon, hydrogen, and oxygen. When a 1.00 g was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were obta

Alina [70]

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

Determine the mass ratio of the constituent elements of the compound.
Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

$\tt \dfrac{12}{44}\times 1.4991=0.409~g$

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

$\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g$

Mass O = Mass sample - (mass C + mass H) :

$\tt 1-(0.409+0.0455)=0.5455~g$

mol ratio C : H : O =

$\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3$

5 0

3 years ago

In order to dry wet clothes, we spread them on a clothline. This is because (i) spreading increases surface area (ii) clothes be

Novosadov [1.4K]

Answer:

this is because spreading it makes more sunlight hit the cloth which results in it drying faster

3 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago

In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account

hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0

3 years ago