What is the solution called in the buret during a titration?

Answers

Answer 1

Answer:

Answer:

The solution in the buret, during a titration is called titrant.

Explanation:

A titration is a useful process, that makes you know the concentration of a solution. A titrant solution (burette) is evaluated against a titrand to control the pH changes against the volume aggregate. Only a strong acid with a strong base, a strong base with a strong acid, a weak acid with a strong base and a weak base with strong acid are valued.

When the pH reaches the equivalence point, it is said that the normality of the acid by the milliequivalents, is equal to the basic normality by its milliequivalents. In conclusion, the entire base / acid became its conjugate pair.

To check this sudden change in pH, a substance is used, called Indicator that changes the color of the titrand (analyte).

Answer 2

Answer:

Final answer:

In a titration analysis, the substance in the buret is called the 'titrant'. It is used to react with the analyte, the sample solution whose concentration we're measuring. The goal is to reach the endpoint, the point when a distinct visual change indicates that the titrant has completely reacted with the analyte.

Explanation:

In a titration analysis, the solution in the buret is called the titrant. This solution contains a known concentration of a substance. During a titration, this titrant is added incrementally to a sample solution, called the analyte, which contains the substance whose concentration is to be measured. The titrant and analyte undergo a chemical reaction of known stoichiometry.

By measuring the volume of the titrant solution needed to completely react with the analyte, scientists can calculate the concentration of the analyte. This point where the titrant has completely reacted with the analyte is termed the equivalence point of the titration. The process of adding the titrant is halted when a distinct change is visually detected in the solution - this could be a color change, for example. This is known as the end point.

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Which term describes this reaction? CH3Br(aq) + OH-(aq) -> CH3OH(aq) + Br-(aq) A. addition B. condensation C. elimination D. substitution

Answers

Answer is: D. substitution.

Substitution reaction or single displacement reaction is a chemical reaction in which one functional group in a chemical compound is replaced by another functional group.

Balanced chemical reaction: CH₃Br(aq) + OH⁻(aq) → CH₃OH(aq) + Br⁻(aq).

In this balanced chemical reaction hydroxy group OH⁻ replaced one atom of bromine (Br⁻) in methyl bromide (CH₃Br).

Answer : The correct option is, (D) Substitution reaction.

Explanation :

Addition reaction : It is a type of reaction in which a molecule combined with the another molecule to give a new larger molecule.

The general representation of this reaction is,

$A+B\rightarrow C$

Condensation reaction : It is a type of reaction in which two molecules combine to form a larger molecule and also producing a smaller molecule such as water.

$R_1-OH+H-OOCR_2\rightarrow R_1-O-COR_2+H_2O$

Elimination reaction : It is a type of reaction in which the loss of elements from the starting material to form a double bond in the product.

The general representation of this reaction is,

$HR_1-R_2X+B\rightarrow R_1=R_2+HB^++X^-$

Substitution reaction : It is a type of reaction in which one functional group is replaced by the another functional group.

The general representation of this reaction is,

$AB+C\rightarrow AC+B$

(B and C are the two different functional groups)

In the given reaction, Br (functional group) is replaced by the hydroxide ion (functional group). So, this reaction is a substitution reaction.

Hence, the given reaction is a substitution reaction.

Which one of the following groups of chemical compounds is composed entirely of organic compounds?A. C2H4O, CH2O, CaSO4
, C3H5
(OH)3
B. C2H2
, CH4
, CaCl2
, CaCN2
C. Ch3OCH3
, Ca3
(PO4
)2
, CO2
, H2CO3
D. C6H6
, C2H5OH, C6H5CH3
, C3H5
(NO3
)3

Answers

Answer;

D. C6H6, C2H5OH, C6H5CH3, and C3H5(NO3)3.

Explanation;

-Organic compound are chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen, like the examples above in choice D. . Inorganic Compoundson the other hand, are compounds made from any elements except those compounds of carbon, such as CaSO4, CaCl2 and Ca3(PO4)2 in A, B and C respectively.

The correct answer is D.

In group A, CaSO4 is not organic; in group B, CaCl2 is not organic; in group C, Ca3(PO4)2 is not organic.

Hope this helps~

Which one of the following statements is correct?A. An electron has a negative electrical charge.
B. An electron has a neutral electrical charge.
C. An electron is much larger than an atom.
D. An electron is located inside the nucleus of an atom.

Answers

Answer: Option (A) is the correct answer.

Explanation:

An atom consists of three sub atomic particles. These particles are protons, electrons and neutrons.

Protons have a positive charge. Whereas neutrons have no charge. On the other hand, electrons have a negative charge.

Thus, we can conclude that out of the given options, an electron has a negative electrical charge is the correct statement.

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answers

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

$\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g$

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL$

(a) Now we have to calculate the volume percent.

$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%$

(b) Now we have to calculate the mass percent.

$\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%$

(c) Now we have to calculate the molarity.

$\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}$

$\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L$

(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}$

$\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg$

(e) Now we have to calculate the mole fraction of ethylene glycol.

$\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}$

$\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole$

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole$

$\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244$

Which of these processes involves the splitting of an atom?nuclear fusionchemical reactionnuclear fissionphysical change

Answers

the process of splitting an atom into two nuclei is the nuclear fission

Given the following reactions 2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJ S (s) + O2 (g) → SO2 (g) ΔH = -297 kJ the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) + O2 (g) → 2SO3 (g) is ________ kJ

Answers

Answer:

-196 kJ

Explanation:

By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.

2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ

S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)

2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ

2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ

-------------------------------------------------------------

2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)

Simplifing the compounds that are in both sides (bolded):

2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ

Final answer:

The enthalpy of the reaction where sulfur dioxide is oxidized to sulfur trioxide is -395 kJ.

Explanation:

The calculation of the enthalpy change of the reaction in which sulfur dioxide is oxidized to sulfur trioxide involves Hess's Law, which states that the enthalpy change of a chemical reaction is the same whether it takes place in one step or several steps. This can be solved by comparing the enthalpy changes given in the two reactions presented.

First, consider the reactions given:

2S(s) + 3O₂(g) → 2SO₃(g), ΔH = -790 kJ

S(s) + O₂(g) → SO₂(g), ΔH = -297 kJ

From these reactions, it is seen that the first reaction can be re-written as:

2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = -790 kJ

However, this reaction contains two moles of SO₂ whereas the reaction in question only requires one mole. Thus, the enthalpy change for the reaction becomes: ΔH = -790 KJ / 2 = -395 kJ.

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