Please help me with question 7. Thank you so much.

Answer:

Bromine will be a liquid

Answer: The volume of NaOH required at the half-equivalence point is 6.21 mL

Explanation:

The chemical equation for the reaction of a weak acid with NaOH follows:

[tex]HA+ NaOH\rightarrow NaA+H_2O[/tex]

From the equation, we can say that NaOH and weak acid is present in a 1 : 1 ratio.

We are given:

Volume of NaOH required at equivalence point = 12.42 mL

The volume of NaOH required at half-equivalence point will be = [tex]\frac{12.42mL}{2}=6.21mL[/tex]

Hence, the volume of NaOH required at the half-equivalence point is 6.21 mL

The volume of NaOH at the half-equivalence point is 6.21 mL

The equivalence point is the point at which equal amount of the acid and base have reacted.

Half equivalence point = Equivalence point / 2

Half equivalence point = 12.42 / 2

Half equivalence point = 6.21 mL

Therefore, we can conclude that the volume of NaOH at the half-equivalence point is 6.21 mL.

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Answer:

the amount of a substance that will dissolve in a given amount of solvent.

Explanation:

Solubility is a term used to describe how readily a substance can be dissolved in a solvent to form a solution. Thus, a substance is said to be soluble if it dissolves completely in a solvent and insoluble if it doesn't dissolve or only dissolves partially.

For example, sodium chloride (NaCl) when mixed with water dissociates into sodium and chloride ions. Thus, salt (sodium chloride) is said to be soluble because it dissolves completely in water.

Furthermore, a compound that dissolves completely in water to produce an aqueous solution is said to be soluble in water.

In conclusion, solubility is simply the amount of a substance such as salt, that will dissolve in a given amount of solvent. A solvent is any liquid such as water, coffee, tea, etc., that dissolves a liquid, gaseous, or solid solute to produce a solution.

Answer:

Concentration C = 0.0189 mol/L

Explanation:

From the given information:

Let consider the formula used in calculating the concentration according to Beer's law:

[tex]\mathtt{A =\varepsilon \times L \times C}[/tex] --- (1)

here;

A = absorbance

ε = coefficient of molar absorptivity

L = path length

C = concentration (mol/L)

Also, from Beer law plot:

y = mx+b

where,

y represent absorbance A

b represents intercept

m represents the coefficient of molar absorptivity ε

and x represents the concentration(C).

replacing the substituted entities

A = ε × C + b ---- (2)

Making the concentration the subject of the above formula:

[tex]C = \dfrac{A-b}{\varepsilon}[/tex]----(3)

From y = 11.589x - 0.0002

A = 11.589 *C - 0.0002

Given that:

A = 0.219

∴

0.219 = 11.589 *C - 0.0002

0.219 + 0.0002 =  11.589 *C

C = 0.2192/11.589

C = 0.0189 mol/L

Answer: The pH of the resulting solution will be 3.60

Explanation:

Molarity is calculated by using the equation:

[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)

We are given:

Molarity of formic acid = 0.100 M

Molarity of potassium formate = 0.100 M

Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]\text{Moles of formic acid}=(0.100mol/L\times 0.420L)=0.0420mol[/tex]

[tex]\text{Moles of potassium formate}=(0.100mol/L\times 0.420L)=0.042mol[/tex]

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

[tex]\text{Moles of KOH}=(1mol/L\times 0.007L)=0.007mol[/tex]

The chemical equation for the reaction of formic acid and KOH follows:

                 [tex]HCOOH+KOH\rightleftharpoons HCOOK+H_2O[/tex]

I:                   0.042     0.007       0.042

C:                -0.007    -0.007     +0.007

E:                  0.035         -           0.049

Volume of solution = [420 + 7] = 427 mL = 0.427 L

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex] .......(2)

Given values:

[tex][HCOOK]=\frac{0.049}{0.427}[/tex]

[tex][HCOOH]=\frac{0.035}{0.427}[/tex]

[tex]pK_a=3.75[/tex]

Putting values in equation 2, we get:

[tex]pH=3.75-\log \frac{(0.049/0.427)}{(0.035/0.427)}\\\\pH=3.75-0.146\\\\pH=3.60[/tex]

Hence, the pH of the resulting solution will be 3.60

Answer:

The answer is cell wall

Explanation:

Because it is

Solution :

  Term                                                                      Best description

Eriochrome Black T                                                          EBT

Starting color, before titration                               Pale purple-pink color

Molecule that has a high affinity for metal ions       Chelating agent

Ethylenediaminetetracetic acid                                    EDTA

Color at the completion of titration                           Blue-gray color    

Answer:

HCl  is a acid

Explanation:

NaOH is base

Nacl is salt

Answer:

Compare the solubility of silver chloride in each of the following aqueous solutions:

a. 0.10 M AgNO3 More soluble than in pure water.

b. 0.10 M NaCI Similar solubility as in pure water

c. 0.10 M KNO3 Less soluble than in pure water.

d. 0.10 M NH4CH3COO

Explanation:

This is based on common ion effect.

According to common ion effect, the solubility of a sparingly soluble salt decreases in a solution containing common ion to it.

The solubility of AgCl(s) is shown below:

[tex]AgCl(s) <=> Ag^{+}(aq)+Cl^-(aq)[/tex]

So, when it is placed in:

a. 0.10 M AgNO3

Due to common ion effect Ag+, its solubility is less in this solution than in pure water.

b. 0.10 M NaCI :

Due to common ion effect Cl-, its solubility is less in this solution than in pure water.

c. 0.10 M KNO3 :

In this solution there is no presence of common ion.

So, the solubility of AgCl in this solution is similar to that of pure water.

d. 0.10 M NH4CH3COO:

In this solution, AgCl forms a precipitate.

So, the solubility of AgCl is more in this solution compared to pure water.

Answer:

Below

Explanation:

Got it right

Check the periodic table, then click electrons

Answer:

In a combination redox reaction, two or more reactants, at least one of which is a(n) element, form a(n) compound. General Reaction: X + Y > Z

In a decomposition redox reaction, a(n) compound forms two or more products, at least one of which is a(n) element. General Reaction: Z>X+Y

In double-displacement (metathesis) reactions, such as precipitation and acid-base reactions, atoms (or ions) of two compounds exchange places; these reactions are not redox processes. General Reaction: AB+CD>AD+CB

In solution, single-displacement reactions occur when a(n) atom of one element displaces the atom of another. Since one of the reactants is a(n) element, all single-displacement reactions are redox processes. General Reaction: X+YZ>XY+Z

Explanation:

In a combination redox reaction, two or more reactants, at least one of which is a(n) element, form a(n) compound.

General Reaction: X + Y > Z

In the reaction scheme above, X combines with Y to give Z as a product.

In a decomposition redox reaction, a(n) compound forms two or more products, at least one of which is a(n) element.

General Reaction: Z>X+Y

In the reaction scheme above, Z decomposes to X and Y

In double-displacement (metathesis) reactions, such as precipitation and acid-base reactions, atoms (or ions) of two compounds exchange places; these reactions are not redox processes since there are no changes occurring in the oxidation number of the atoms (or ions) involved.

General Reaction: AB+CD>AD+CB

In the reaction scheme above, B and D exchange places in their respective compounds

In solution, single-displacement reactions occur when a(n) atom of one element displaces the atom of another. This type of reaction is due to the difference in the reactivities of the elements. The more reactive atom of one element displaces the least reactive atom of another element from its solution.

Since one of the reactants is a(n) element, all single-displacement reactions are redox processes.

General Reaction: X+YZ>XY+Z

In the reaction scheme above, X displaces Z from the compound YZ.

Answer:

1.75L

Explanation:

Reaction of decomposition is:

2KClO₃(s) →  2KCl(s) + 3O₂(g)

We determine moles of salt:

4.35 g . 1 mol /122.55 g = 0.0355 moles

Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen

Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.

We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:

We convert T° from °C to K → 27°C + 273 = 300K

P . V = n . R . T

0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K

V = (0.0532 mol  . 0.0821 L.atm/mol.K . 300K) / 0.75 atm

V  = 1.75 Liters

Answer:

the TWO characteristics of an effective collision are:

1.Molecules collide with sufficient energy

2.Molecules collide with the proper orientation.

Answer:

5.37 × 10⁻⁴ mol/L

Explanation:

A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Step 1: Given data

Step 2: Calculate the concentration of the final solution

We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L

Answer:

V = 365.54 mL

Explanation:

Given that,

The density of chloroform, d = 1.48 g/mL

The mass of chloroform, m = 541 g

We need to find the volume of chloroform.

We know that,

Density = mass/volume

So,

[tex]V=\dfrac{m}{d}\\\\V=\dfrac{541\ g}{1.48\ g/mL}\\\\=365.54\ mL[/tex]

So, the volume of chloroform is 365.54 mL.

Answer:

Wide melting point range - impure sample with multiple compounds

Experimental melting point is close to literature value - pure sample of a single compound

Experimental melting point is below literature value - impure sample with multiple compounds

Narrow melting point range - pure sample of a single compound

Explanation:

The melting point of substances are easily obtainable from literature such as the CRC Handbook of Physics and Chemistry.

A single pure substance is always observed to melt within a narrow temperature range. This melting temperature is always very close to the melting point recorded in literature for the pure compound.

However, an impure sample with multiple compounds will melt over a wide temperature range. We also have to recall that impurities lower the melting point of a pure substance. Hence, the experimental melting point of an impure sample with multiple compounds is always below the literature value.

Answer:

sim eu também preciso desta respota

An ether and alkene are formed as by products in the reaction which is a electrophilic addition reaction.

An addition reaction known as an electrophilic addition reaction occurs when a chemical molecule having a double or triple bond has one of its bonds broken and two new bonds are formed. The interconversion of C=C and CC into a variety of significant functional groups, such as alkyl halides and alcohols, is made possible via the key.

The following describes the general mechanism: Hydrogen bromide produces an electrophile, H+, which attacks the double bond to create a carbocation. The production of ions is dominated by secondary carbocation because it is more stable than primary carbocation.

Thus, an ether and alkene are formed as by products in the reaction which is a electrophilic addition reaction.

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Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

Answer:

No, the correct empirical formula is [tex]C_3H_7[/tex].

Explanation:

Hello there!

In this case, according to the given information, it turns out necessary for us to bear to mind the fact that empirical formulas must not be expressed in decimal numbers, for that reason, we need to multiply the given empirical formula by 2 to get the correct one:

[tex]C_3H_7[/tex]

Which is now possible.

Regards!

Answer:

soy....plant protein also is found in vegtables and grain products.

The protein that is derived from the plant is soyabean. Protein-rich foods have a rich source of amino acids. Amino acids are required by the body for different cellular activities.

Proteins are present in plant-based foods and animal-based foods. In nature, the plant-based foods that are rich in proteins are soybeans, beans etc. Animal products such as dairy products, milk, and cheese are rich sources of proteins.

Animal-based foods such as eggs, meat, and fish have a good amount of protein. The proteins are made up of amino acids. Peptide bonds connect each amino acid to the next.

Proteins are necessary for the body, as cells need amino acids for both structural and functional support. Not all cells of the body synthesise all amino acids. Some essential amino acids are required by the body and can be taken from external sources such as plants.

Hence, the plant-based protein is soyabean.

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The question is incomplete; the complete question may be the following:

1) Which of the following is a protein source of plant origin?

A)egg

B) Soyabean

C)cheese

D)Milk

Answer:

Amount of reactant after four hours = 4,26 grams

Explanation:

Suppose y denotes the amount of reactant at the time (t)

The given function:

[tex]\dfrac{dy}{dt} = -0.7 y[/tex]

[tex]\dfrac{dy}{y} = -0.7 dt[/tex]

Taking integral on both sides

㏑(y) = -0.7t + c¹

[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]

[tex]y(t) = Ce ^{-0.7t}[/tex]

At t = 0 ; y (t) = 70

∴

[tex]70 = Ce^{-0.7(0)}[/tex]

C = 70

As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]

After four hours, the amount of the reactant is:

[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]

[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]

[tex]\mathtt{y(t) = 4.26}[/tex]

Amount of reactant after four hours = 4,26 grams

Answer:

Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.

When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.

Explanation:

Constitutional isomers are the one which differs in the structural formula.

When compound A is treated with sodium methoxide, a substitution reaction predominates.

That means sodium methoxide is a strong base and a strong nucleophile.

But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.

The reaction and the structures of A and B are shown below:

Explanation:

can u post a picture of the question ?

Answer:

(E)-1-phenylbut-1-ene

Explanation:

2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is obtained from the sequence of the reactions of the 2-phenyl-4 bomobutane when it reacts with [tex]NaN_3[/tex] is the (E)-1-phenylbut-1-ene.

Thus the answer is 2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is (E)-1-phenylbut-1-ene.

Answer:

See Explanation

Explanation:

Rate of reaction refers to how quickly or slowly a reaction proceeds. The rate of reaction depends on certain factors.

Two among the factors that affect the rate of reaction are the concentration of reactants and the surface area of reactants.

The more the concentration of reactants, the faster the rate of reactants because there is a high possibility of collision between reactants. Also, the higher the surface area of reactants, the greater the rate of reaction.

In flour mills and coal mines where there is a large amount (concentration) of combustible materials and the particles are powdered (high surface area), there is a greater risk of explosion due to a high rate of reaction owing to a combination of the two factors discussed above.

Answer:

0.186 moles

Explanation:

In order to convert grams of PCl₅ into moles, we need to use its molar mass:

Then we proceed to calculate the number of moles:

In the given question Phosphorus pentachloride is used as a catalyst for certain chemical reaction. 38.7 g of [tex]\rm PCl_5[/tex], there are 0.186 moles of [tex]\rm PCl_5[/tex].

A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy required for the reaction to occur.

To calculate the number of moles in 38.7 g of [tex]\rm PCl_5[/tex], we need to divide the given mass of [tex]\rm PCl_5[/tex] by its molar mass.

The molar mass of [tex]\rm PCl_5[/tex] can be calculated by adding the atomic masses of one phosphorus atom and five chlorine atoms:

Molar mass of  [tex]\rm PCl_5[/tex] = (1 x atomic mass of P) + (5 x atomic mass of Cl)

= (1 x 30.97 g/mol) + (5 x 35.45 g/mol)

= 208.22 g/mol

Now, we can calculate the number of moles of  [tex]\rm PCl_5[/tex]:

Number of moles = mass / molar mass

= 38.7 g / 208.22 g/mol

= 0.186 moles

Therefore, there are 0.186 moles of  [tex]\rm PCl_5[/tex] in 38.7 g of  [tex]\rm PCl_5[/tex].

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Answer:

yo its jess bregoli

your answer is given below

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An atomic mass unit is defined as a mass equal to one twelfth the mass of an atom of carbon-12. The mass of any isotope of any element is expressed in relation to the carbon-12 standard. For example, one atom of helium-4 has a mass of 4.0026 amu. An atom of sulfur-32 has a mass of 31.972 amu.

Answer:

Exothermic

1771 g

Explanation:

Step 1: Write the balanced thermochemical equation

2 S(s) + 3 O₂(g) ⇒ 2 SO₃(g)   ΔH° = -791.4 kJ

Since ΔH° < 0, the reaction is exothermic.

Step 2: Calculate the moles of SO₃ produced when 8753 kJ of energy are released

According to the thermochemical equation, -791.4 kJ are released every 2 moles of SO₃ that are formed.

-8753 kJ × 2 mol/(-791.4 kJ) = 22.12 mol

Step 3: Calculate the mass corresponding to 22.12 moles of SO₃

The molar mass of SO₃ is 80.06 g/mol.

22.12 mol × 80.06 g/mol = 1771 g