why is a chemical indicator added to the flask in which a titration will occur?

Answer: The molar mass of solute is 156 g/mol

Explanation:

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

The equation used to calculate elevation in boiling point follows:

[tex]\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}[/tex]

[tex]\Delta T_b[/tex] = ? °C

Boiling point of pure water = 100°C

Boiling point of solution = 101°C  

Putting values in above equation, we get:

[tex]\Delta T_b=(101-100)^oC=1^oC[/tex]

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

[tex]\Delta T_b=iK_bm[/tex]

Or,

[tex]\Delta T_b=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]

where,

[tex]\Delta T_b[/tex] = 1°C

i = Vant hoff factor = 1 (For non-electrolytes)

[tex]K_b[/tex] = molal boiling point elevation constant = 0.52°C/m.g

[tex]m_{solute}[/tex] = Given mass of solute = 300 g

[tex]M_{solute}[/tex] = Molar mass of solute  = ?

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 1000 g

Putting values in above equation, we get:

[tex]1^oC=1\times 0.52^oC/m\times \frac{300\times 1000}{M_{solute}\times 1000}\\\\M_{solute}=156g/mol[/tex]

Hence, the molar mass of solute is 156 g/mol

Answer: 55.35ml

Explanation:

[tex]Ca+2HF\rightarrow CaF_2+H_2[/tex]

To calculate the given moles, we use the formula:

[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]Moles=\frac{85.8g}{78g/mol}=1.1moles[/tex]

By  stoichiometry of the reaction,

1 mole of [tex]CaF_2[/tex]  is produced by 1 mole of Ca

1.1 moles of [tex]CaF_2[/tex]  are produced by=[tex]\frac{1}{1}\times 1.1=1.1 moles[/tex] of Ca.

Mass of [tex]CaF_2=\text{no of moles}\times \text{Molar mass}[/tex]

Mass of [tex]CaF_2=1.1\times {78}=85.8g[/tex]

[tex]Volume=\frac{Mass}{Density}[/tex]

[tex]Volume=\frac{85.8g}{1.55g/ml}=55.35ml[/tex]

Final answer:

To calculate the mass percentages of Cl and Br in a mixture of AgCl and AgBr with 60.94% Ag by mass, we need additional data to determine the individual masses of Cl and Br. Without it, we cannot provide precise percentages.

Explanation:

The question is about calculating the mass percents of chlorine (Cl) and bromine (Br) in a mixture of silver chloride (AgCl) and silver bromide (AgBr) that contains 60.94% silver (Ag) by mass. The molar mass of silver (Ag) is 107.87 g/mol, for AgCl it is 143.32 g/mol (107.87 g/mol + 35.45 g/mol for Cl), and for AgBr it is 187.77 g/mol (107.87 g/mol + 79.90 g/mol for Br).

To find the mass percentages of Cl and Br in the mixture, we will first assume we have 100 grams of the mixture. Given the percent of silver, we can calculate the mass of silver in the mixture, which should be 60.94 grams. The remaining mass will be the combined mass of Cl and Br. However, without additional information such as the mass or molar ratios of AgCl to AgBr, we cannot precisely divide the remaining mass between Cl and Br. If such data were available, the mass percentages of Cl and Br could be found by calculating their individual masses and dividing by the total mass of the mixture, then multiplying by 100.

[tex]\boxed{1.1255 \times {{10}^{23}}{\text{ atoms}}}[/tex] of oxygen are present in 7.00 g of sodium dichromate.

Further Explanation:

Given information:

Mass of sodium dichromate: 7.00 g

To calculate:

Number of oxygen atoms in 7.00 g of sodium dichromate

Steps to proceed:

I. First of all, moles of sodium dichromate that are present in 7.00 g of sodium dichromate are to be calculated. This is done with the help of equation (1) as mentioned below.

The formula to calculate moles of sodium dichromate [tex]\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)[/tex] is as follows:

[tex]{\text{Moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} = \dfrac{{{\text{Mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}}{{{\text{Molar mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}}[/tex]                                               …… (1)

The mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}[/tex] is 7.00 g.

The molar mass of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}[/tex] is 261.97 g/mol.

Substitute these values in equation (1).

[tex]\begin{aligned}{\text{Moles of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}} &= \frac{{{\text{7}}{\text{.00 g}}}}{{{\text{261}}{\text{.97 g/mol}}}} \\ &= 0.0267{\text{ mol}} \\ \end{aligned}[/tex]  

II.Moles of oxygen that are present in 0.267 moles of sodium dichromate are to be calculated.

According to the chemical formula of sodium dichromate, it is evident that one mole of sodium dichromate contains two moles of sodium, two moles of chromium and seven moles of oxygen in it. Since one mole of sodium dichromate consists of seven moles of oxygen, moles of oxygen present in 0.267 moles of sodium dichromate can be calculated as follows:

[tex]\begin{aligned}{\text{Moles of oxygen}} &= \left( {0.0267{\text{ mol N}}{{\text{a}}_2}{\text{C}}{{\text{r}}_4}{{\text{O}}_7}} \right)\left( {\frac{{7{\text{ mol O}}}}{{{\text{1 mol N}}{{\text{a}}_2}{\text{C}}{{\text{r}}_4}{{\text{O}}_7}}}} \right) \\&= 0.1869{\text{ mol O}} \\\end{aligned}[/tex]  

III. Number of oxygen atoms that are present in 0.1869 moles of oxygen is to be calculated. This is done with the help of Avogadro’s law which states that one mole of a substance contains [tex]6.022 \times {10^{23}}{\text{ particles}}[/tex] . Such particles can be atoms, molecules, or formula units.

Since one mole of oxygen contains [tex]6.022 \times {10^{23}}{\text{ atoms}}[/tex], atoms of oxygen present in 0.1869 moles of oxygen can be evaluated as follows:

[tex]\begin{aligned}{\text{Number of oxygen atoms}} &= \left( {0.1869{\text{ mol}}} \right)\left( {\frac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{{\text{1 mol}}}}} \right) \\&= 1.1255 \times {10^{23}}{\text{ atoms}} \\\end{aligned}[/tex]  

Hence, [tex]1.1255 \times {10^{23}}{\text{ atoms}}[/tex] of oxygen are present in 7.00 g of sodium dichromate.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: sodium dichromate, oxygen, mass, molar mass, 7.00 g, 1.1255*10^23 atoms, Avogadro’s law, one mole, atoms, molecules, particles.

The heat change for the formation of 1 mol of aluminum oxide from its elements is 1676 kJ.

The heat change for the formation of 1 mol of aluminum oxide from its elements can be calculated using the information provided in the question. We know from the given thermochemical equation that the decomposition of 2 mol of aluminum oxide releases 3352 kJ of heat. Therefore, we can set up a proportion to find the heat change for the formation of 1 mol of aluminum oxide:

3352 kJ / 2 mol = x kJ / 1 mol

Solving for x gives us the heat change for the formation of 1 mol of aluminum oxide, which is 1676 kJ.

The balanced equation is :

2Al + 3CuSO4 -> 3Cu + Al2(SO4)3.

A single-displacement reaction takes place while an element replaces any other element in a compound. A metallic handiest replaces steel, and a nonmetal only replaces a nonmetal. only a greater reactive element can update the opposite element within the compound with which it reacts.

A single substitute response, now and again known as a single displacement response, is a reaction in which one element is substituted for another detail in a compound. The starting materials are continually natural elements, which include pure zinc steel or hydrogen fuel, plus an aqueous compound.

To determine whether or not a given single replacement will occur, you have to use an “activity series” table. If the metal or the halogen is above the detail it will replace based totally on the hobby series, a single displacement response will occur. Examples of displacement reactions are The reaction among iron and copper sulphate to present iron sulphate as a product. here, iron displaces copper due to the fact iron is extra reactive than copper. The response between zinc and iron sulphate to offer zinc sulphate as a product.

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To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

pH = - log [H+]            where H+ must be in Molar

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2] 

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 
[OH-] = 10^-14 / [H+]

[OH-] = 10^-14 / 3.25 × 10-2

[OH-] = 3.08 × 10-13 M

Answers:

Acidic

[OH-] = 3.08 × 10-13 M

The hydrogen ion concentration is 3.25 × 10-2 M, indicating an acidic solution with a pH of 1.49. The hydroxide concentration is 3.08 × 10-13 M.

In this question, the hydrogen ion concentration is given as 3.25 × 10-2 M. This concentration indicates that the solution is acidic, with a pH value equal to -log[H3O+]. Therefore, pH = -log(3.25 × 10-2) = 1.49.

The hydroxide concentration can be calculated using the formula: [OH-] = Kw/[H3O+]. At 25°C, the value of Kw (the ion-product constant for water) is 1.0 × 10-14. Therefore, [OH-] = 1.0 × 10-14 / 3.25 × 10-2 = 3.08 × 10-13 M.

We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) 
Ao = amount at time zero = 0.00560 M 
k = rate constant
t = time = 1.50 hours or 5400 s 

The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

Plugging in the values in the 1st equation:

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 × 10^–4 M           (simplify as necessary)

To calculate the concentration of cyclopropane that remains after 1.50 hours, we need to use the first-order integrated rate law equation. Without the rate constant, we cannot calculate the exact concentration, but we can determine that it will be less than 0.200 mol/L.

To calculate the concentration of cyclopropane that remains after 1.50 hours, we need to use the first-order integrated rate law. The rate constant for the reaction is not given, so we cannot calculate it. However, we can use the given information that at 10.0 minutes (0.1667 hours) the concentration of cyclopropane is 0.200 mol/L and find the concentration at 1.50 hours using the integrated rate law equation:

[tex][C] = [C]0 * e^(-kt)[/tex]

where [C] is the concentration at time t, [C]0 is the initial concentration, k is the rate constant, and t is the time.

Given [C]0 = 0.200 mol/L, t = 1.50 hours, and [C] = ?

Let's solve the integrated rate law equation:

Plug in the known values:

[tex][C] = 0.200 * e^(-k*1.50)[/tex]

Since the rate constant is not given, we cannot calculate the exact concentration. However, we can still make a general qualitative statement that the concentration will be less than 0.200 mol/L.

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  CS2(l)  +Cl2(g)→CCl4(l)   + S2Cl2 (l)

The next  logical  step  in balancing  the  equation above is  place  the  coefficient  3  in front  of the chlorine  molecule  ( answer B)

  Explanation

When   coefficient 3  in placed in front  of   chlorine  molecule  the balanced equation is as below

The compound CH3-CH2-SH belongs to the organic family known as Thiols. Thiols are structurally similar to alcohols but the oxygen atom in alcohols is replaced by sulfur in thiols. Both thiols and alcohols belong to a larger class called the Hydrocarbon Derivatives.

The compound you mentioned, CH3-CH2-SH, falls into the organic family categorized as 'Thiols'. Thiols are similar to alcohols in structure, but they have a sulfur atom rather than an oxygen atom. They are identified by the -SH or sulfhydryl functional group. Like the hydroxyl group in an alcohol, this sulfhydryl group gives a thiol its properties.

For example, the compound CH3-SH, also known as 'Methanethiol', is analogous to the alcohol methanol. Both thiols and alcohols are part of a larger class called the 'Hydrocarbon Derivatives'. As the name suggests, these are derived from hydrocarbons by replacing one or more hydrogen atoms with a functional group, which in this case is either a hydroxyl group (-OH) for alcohols or a sulfhydryl group (-SH) for thiols.

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Answer:The specific heat of the unknown substance is 1.22 J/ °Cg.

Explanation:

Heat absorbed by the water .

Mass of the water = 110.0 g

Change in temperature of water = [tex]\Delta T=32.4^oC-24.2^oC=8.2^oC[/tex]

[tex]Q=mc\Delta T=110.0g\times 4.18 J/^oCg\times 8.2^oC=3,770.36 J[/tex]

Heat lost by substance(Q') = Heat gained by the water(Q)

-Q' = Q

Change in temperature of the substance =

[tex]\Delta T'=(32.4^oC)-105.0^oC=-72.6 ^oC[/tex]

Mass of the substance = m'=42.5 g

Specific heat of substance = c'

[tex]-(m'\times c'\times \Delta T')=3,770.36 J[/tex]

[tex]c'=\frac{3,770.36 J}{42.5 g\times 72.6^oC}=1.22 J/^oCg[/tex]

The specific heat of the unknown substance is 1.22 J/ °Cg.

Answer:

The correct formula for the compound made when aqueous solutions containing a dissolved Magnesium and Chloride are mixed is Magnesium chloride [tex]MgCl_2[/tex]

Explanation:

If we suppose the existence of two components in aqueous solutions containing Magnesium and Chloride. For example: Hydrochloric acid - HCl and Magnesium hydroxide - [tex]Mg(OH)_2[/tex].

The global reaction would be:  

HCl + [tex]Mg(OH)_2[/tex] --> [tex]MgCl_2[/tex] + [tex]H_2 O[/tex]

It means, Hydrochloric acid is neutralized with Magnesium hydroxide to produce Magnesium chloride and Water.  

Besides of that, we can analyze the aqueos solution of every componenent:

For Hydrochloric acid:

HCl + [tex]H_2 O[/tex] --> [tex]H_3O^+[/tex] + [tex]Cl^-[/tex]

For Magnesium hydroxide

[tex]Mg(OH)_2[/tex] +  [tex]H_2 O[/tex] --> [tex]H_2O-(OH)^-[/tex] + [tex]Mg^+2[/tex]

Finally, the ionic compounds will form the salt:

[tex]Mg^+2[/tex] + [tex]Cl^-[/tex] ->  [tex]MgCl_2[/tex]

Final answer:

Baking soda, or sodium bicarbonate, when dissolved in water, has a basic pH close to 9.0, classifying it as a weak base.

Explanation:

When sodium bicarbonate, commonly known as baking soda, is dissolved in water, it exhibits basic properties. The pH of a baking soda solution is closest to 9.0, which means it is a weak base. Water is neutral with a pH of 7.0, whereas a substance with a pH higher than 7.0 is considered alkaline or basic.

The pH scale ranges from 0 to 14, with 7 being neutral. Acids have a pH less than 7, and bases have a pH greater than 7. The pH of a substance is determined by its concentration of hydronium ions. Baking soda has a lower concentration of hydronium ions compared to pure water, thereby making it a base with a pH higher than 7.

In practical applications, baking soda is used to neutralize acids, such as in bee sting treatments or when mixed with acetic acid in vinegar during bread preparation.

Answer : The molar concentration of [tex]K^+[/tex] ion is, 0.5 M

Explanation :

The dissociation reaction of [tex]K_2SO_4[/tex] is,

[tex]K_2SO_4(aq)\rightarrow 2K^+(aq)+SO_4^{2-}(aq)[/tex]

By the stoichiometry we can say that, 1 mole of [tex]K_2SO_4[/tex] dissociates into 2 mole of [tex]K^+[/tex] ions and 1 mole of [tex]SO_4^{2-}[/tex] ions.

As we  are given the concentration of [tex]K_2SO_4[/tex] is, 0.250 M.

So, the molar concentration of [tex]K^+[/tex] ion = [tex]2\times 0.250=0.5M[/tex]

The molar concentration of [tex]SO_4^{2-}[/tex] ion = 0.250 M

Therefore, the molar concentration of [tex]K^+[/tex] ion is, 0.5 M

To calculate the pH of a 1.60 M acetic acid solution, use the ionization constant (Ka) of acetic acid, set up an ICE table, and solve for x to find the concentration of the hydronium ion (H3O+). Then, calculate the pH using the equation pH = -log[H3O+]. The pH of the acetic acid solution is approximately 1.5873.

To calculate the pH of a 1.60 M acetic acid solution, we can use the ionization constant (Ka) of acetic acid and the relationship between the concentration of the acid, its conjugate base, and the pH. The Ka value for acetic acid is 1.8×10-5. Since acetic acid is a weak acid, we can assume that the concentration of the hydronium ion (H3O+) is equal to the concentration of the acid that has ionized.

Using the equation Ka = [H3O+][C2H3O2]/[HC2H3O2], we can set up an ICE table to calculate the concentration of the hydronium ion:

Substituting these values into the Ka expression, we get 1.8×10-5 = (1.60 - x)(x)/(1.60 - x). Solving for x gives us x ≈ 0.0127 M. Since [H3O+] = 1.60 - x, the pH of the solution is approximately 1.60 - 0.0127 = 1.5873.

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The pH of a 1.60 M acetic acid solution can be calculated using the equilibrium constant expression and initial concentration. The pH is equal to -log[H3O+]. The pH of this particular solution is 2.37.

The pH of a 1.60 M acetic acid solution can be calculated using the equation:

pH = -log[H3O+]

First, we need to find the concentration of the hydronium ion, [H3O+]. To do this, we use the equilibrium constant expression for acetic acid, which is Ka = [H3O+][CH3CO2-] / [CH3COOH]. With a given Ka value of 1.8 × 10-5 and a concentration of acetic acid of 1.60 M, we can plug in these values to find the concentration of [H3O+], which is then used to calculate the pH.

Therefore, the pH of a 1.60 M acetic acid solution is 2.37.

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Explanation:

The given data is as follows.

        Molarity of solution = 0.184 M

        Volume of solution = ?

         number of moles = 0.113 mol

As molarity is the number of moles present in liter of solvent.

Mathematically,      Molarity = [tex]\frac{\text{No. of moles}}{\text{Volume}}[/tex]

Hence, calculate the volume of given solution as follows.

                  Molarity = [tex]\frac{\text{No. of moles}}{\text{Volume}}[/tex]

                  0.184 M = [tex]\frac{0.113 mol}{volume}[/tex]

                     volume = 0.614 L

As 1 L = 1000 mililiter. Hence convert 0.614 L into ml as follows.

                      [tex]0.614 L \times \frac{1000 ml}{1 L}[/tex]

                       = 614 ml

Thus, we can conclude that the volume of given solution is 614 ml.

To solve this we use the equation, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

40% x 30 L = 30% x V2

V2 = 40 L 

Therefore, you will need to have 30 mL of the 40% acid solution and 10 mL of distilled water. In mixing the two liquids, you should remember that the order of mixing would be acid to water. So, you use a 40 mL volumetric flask . Put small amount of distilled water and add the 30 mL of HCl solution. Lastly, dilute with distilled water up to the mark.

Mole is equal to the Avogadro's constant. The 3.675 moles of NO is required to completely react with 2.45 mole of Ammonia.    

Mole:

One mole is the unit of measurement of small particles like ions, atoms , and molecules. It is equal to the Avogadro's constant.

Given reaction,

[tex]\bold{4 NH_3 + 6 NO \rightarrow5 N_2 + 6 H_2O}[/tex]

Means 4 moles of Ammonia react with 6 moles of NO in reaction. the molar ratio of ammonia and NO is 4:6.

Hence,

Moles of NO required

[tex]\Rightarrow\bold{\frac{6}{4} \times 2.45 = 3.675 mole}[/tex]          

Hence, we can conclude that the 3.675 moles of NO is required to completely react with 2.45 mole of Ammonia.    

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Scientific theories become stronger over time primarily because scientists continuous attempt to challenge and refine them, which leads to improvements and adjustments that further solidify the theory.

The best explanation for why scientific theories grow stronger over time is option A: 'As more scientist attempt to find holes in the current theory adjustments are made which make the theory stronger.' Scientific theories are rigorously tested and often challenged by researchers who seek to uncover any weaknesses or inconsistencies. When such efforts are made, they can lead to refinements and adjustments that subsequently strengthen the theory. This process may involve incorporating new evidence, reinterpreting existing data, or modifying theoretical frameworks to better explain the phenomenon in question.

Scientific theory is not just an educated guess; it is a comprehensive explanation supported by many research studies that collectively provide falsifiable evidence. Throughout history, theories such as the heliocentric model have been revised and strengthened through this process. It is by this meticulous scrutiny, including the elimination of competing hypotheses and the aggregation of reliable evidence, that a scientific theory matures and gains acceptance.

The statement which best explains why scientific theories grow stronger over time is A.) As more scientists attempt to find holes in the current theory, adjustments are made which make the theory stronger.

Scientific theories are dynamic and grow stronger through a process of rigorous testing, criticism, and refinement. When scientists develop a theory, they propose explanations for phenomena observed in the natural world. These theories are not merely guesses; they are grounded in substantial evidence that has been gathered through experimentation and observation. The process of scientific investigation includes trying to refute or find the limitations of current theories. As researchers explore these theories in different contexts and conditions, they either gather more evidence that supports the theory or uncover discrepancies that prompt modifications. This critical assessment and ongoing peer review contribute to the strength and reliability of scientific theories over time. Each time a theory withstands scrutiny or is adjusted to incorporate new data, it becomes more robust and dependable. Scientific theories are considered strong because they have been tested and falsified repeatedly, not because they are simply old or widely taught. It is the accumulation of evidence and the continuous critical analysis by the scientific community that gradually builds the strength of a theory.

At the stoichiometric point of the titration of acetic acid with a strong base, the species present in the solution are acetate ions (CH3COO-) and water (H2O).

During the titration of acetic acid (CH3COOH) with a strong base (NaOH), at the stoichiometric point all of the acetic acid will have reacted with the sodium hydroxide. This means that the species present in the solution at the stoichiometric point are the acetate ions (CH3COO-) and water (H2O). The balanced chemical equation for the reaction is:

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)