Consider that a sample of a compound is decomposed and the masses of its constituent elements is as follows: 1.443 g Se, 0.5848 g O What would be the empirical formula for this compound?

Answer:

The answer is 5.7 minutes

Explanation:

A first-order reaction follow the law of [tex]Ln [A] = -k.t + Ln [A]_{0}[/tex]. Where [A] is the concentration of the reactant at any t time of the reaction, [tex][A]_{0}[/tex] is the concentration of the reactant at the beginning of the reaction and k is the rate constant.

Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be [tex][A]=\frac{6.25}{100}.[A]_{0}[/tex]. And the rate constant (k) is 8.10×10−3 s−1

Replacing the equation of the law:

[tex]Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}[/tex]

Clearing the equation:

[tex]Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t[/tex]

Considering the property of logarithms: [tex]Ln A - Ln B = Ln \frac{A}{B}[/tex]

Using the property:

[tex]Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t[/tex]

Clearing t and solving:

[tex]t = \frac{Ln \frac{6.25}{100} }{-8.10x10^{-3}s^{-1} } = 342.3s[/tex]

The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:

[tex]342.3x\frac{1min}{60s} = 5.7min[/tex]

Answer:

The reaction of 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] can produce 81 g of Water.    

Explanation:

We have the reaction:

[tex]2B_5H_9 + 12O_2 -> 5B_2O_3 + 9H_2O[/tex]

There are 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex].  

First step to calculate the water produced is to find the limit reagent. Molecular weights for those substances involved in the chemical reaction are:  

[tex]B_5H_9[/tex] = 63.12 g/mol

[tex]O_2[/tex] = 32 g/mol

[tex]B_5O_3[/tex] = 69.62 g/mol

[tex]H_2O[/tex] = 18 g/mol

Now, we can express 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] as mol dividing the mass using the molecular weight.

[tex]192 g O_2 (\frac{1mol O_2}{32 g O_2} )= 6 mol O2\\ 126 g B_5H_9 (\frac{1mol B_5H_9}{63.12 g B_5H_9} ) = 1.99 mol B_5H_9[/tex]

After that, we should divide the mol of reagent with their respective stoichiometric coefficient to find the limit reagent so:

[tex]6 mol O_2 / 12 mol O_2 = 0.5 \\1.99 mol B_5H_9 /2 mol B_5H_9 = 0.99 \\[/tex]

It means the limit reagent is Oxygen - [tex]O_2[/tex].

Second step to calculate water produced is to use stoichiometric calculations using oxygen amount. According to the balanced equation 12 mols of [tex]O_2[/tex] will produce 9 mols of [tex]H_2O[/tex] . After that, using molecular weight of water [tex]H_2O[/tex] = 18 g/mol we can calculate the mass. It is shown in the next equations:  

[tex]6 molO_2 (\frac{9molH_2O}{12molO_2} ) (\frac{18 gH_2O}{1molH_2O} ) = 81 g H_2O[/tex].

Finally, we found that the reaction of 192 g of [tex]O_2[/tex] and 126 g of [tex]B_5H_9[/tex] can produce 81 g of Water.  

Using stoichiometry and information from the balanced chemical equation, we can determine that the reaction of 126 g B5H9 and 192 g O2 can produce approximately 162 g of water.

This question is about a chemical reaction between B5H9 and O2, producing water (H2O), among other substances. To figure out how much water can be produced, we need to understand stoichiometric calculations. Analyzing the given balanced equation, we can see that 2 moles of B5H9 react with 12 moles of O2 to produce 9 moles of water.

To find the quantity of water produced (in grams), we first need to find out how many moles of both B5H9 and O2 we have. The molar mass of B5H9 is around 63 g/mol, and for O2 it's 32 g/mol. So, we have approximately 2 moles of B5H9 and 6 moles of O2. The limiting reagent in this case is B5H9 since it will finish first in the chemical reaction.

As per the balanced equation, 2 moles of B5H9 can produce 9 moles of water. So, 2 moles of our limiting reagent (B5H9) will produce 9 moles of water. Since the molar mass of water is 18 g/mol, we multiply 9 moles by 18 g/mol to get approximately 162 g of water that can be produced.

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Answer:

C7H5N3O6

Explanation:

C = 12 g/mol

O = 16 g/mol

H = 1 g/mol

N = 14 g/mol

CO2 = 44 g/mol

H2O = 18 g/mol

NH3 = 17 g/mol

* 44 g CO2 is produced from 12 g C

0.213 g CO2 is produced from = (12)(0.213) / 44 = 0.0581 g C present in 0.157 g compound.

* 18 g H2O is produced from 2 g H

0.0310 g H2O is produced from = (2)(0.0310) / 18 = 0.0034 g H present in 0.157 g compound.

* 17 g NH3 is produced from 14 g N

0.0230 g NH3 is produced from = (14)(0.0230) / 17 = 0.0189 g N present in 0.103 g compound.(this is different)

If in 0.103 g compound there is 0.0189 g N

in 0.157 g compound there is ;

(0.157)(0.0189) / 0.103 = 0.0288 g N

* Mass of C, H and N = 0.0581 + 0.0034 + 0.0288 = 0.0903 g

The rest is mass of O = 0.157 - 0.0903 = 0.0667 g

Moles of elements in the compound sample:

C = 0.0581 g / 12 g/mol = 0.00484 mol

H = 0.0034 g / 1 g/mol = 0.0034 mol

N = 0.0288 g / 14 g/mol = 0.00205 mol

O = 0.0667 g / 16 g/mol = 0.00417 mol

Smallest mole ratio:

N = 0.00205 / 0.00205 = 1.00

C = 0.00484 / 0.00205 = 2.36

H = 0.0034 / 0.00205 = 1.66

O = 0.00417 / 0.00205 = 2.00

Multiplying by 3 we get;

N = 3

C = 7

H = 5

O = 6

Simplest formula:

C7H5O6N3

This is probanly TNT (trinitrotoluen)

Final answer:

The empirical formula of the compound is determined by converting the masses of CO2, H2O, and NH3 to moles of C, H, and N, respectively, and then finding the simplest whole number ratio of these elements.

Explanation:

To determine the empirical formula of the compound, we must first find the number of moles of carbon, hydrogen, and nitrogen in the given samples.

From the combustion data of carbon to CO₂, we use the mass of CO₂ to find the moles of carbon. The molar mass of CO₂ is 44.01 g/mol, so:

0.213 g CO₂ × (1 mol CO₂ / 44.01 g CO₂) = 0.00484 mol CO₂

Each mole of CO₂ contains 1 mole of carbon, so:

0.00484 mol CO₂ × (1 mol C / 1 mol CO₂) = 0.00484 mol C

Similarly, from the mass of H₂O, we find the moles of hydrogen. The molar mass of H₂O is 18.02 g/mol, so:

0.0310 g H₂O × (1 mol H₂O / 18.02 g H₂O) = 0.00172 mol H₂O

Each mole of H₂O contains 2 moles of hydrogen, hence:

0.00172 mol H₂O × (2 mol H / 1 mol H₂O) = 0.00344 mol H

For nitrogen, the molar mass of NH₃ is 17.03 g/mol, which means:

0.0230 g NH₃ × (1 mol NH₃ / 17.03 g NH₃) = 0.00135 mol NH₃

Each mole of NH₃ contains 1 mole of nitrogen, thus:

0.00135 mol NH₃ × (1 mol N / 1 mol NH₃) = 0.00135 mol N

Finally, to find the ratio of atoms in the empirical formula, divide the moles of each element by the smallest number of moles obtained:

The empirical formula is C₄H₃N. The oxygen content is assumed to make up the remainder of the compound's mass. However, since the combustion analysis provides information for C, H, and N only, we'll stop at determining their ratios.

Answer:

The names or formulas are below, just look for the the names or symbols in the tables, write them and cross the powers without the signs.

Explanation:

Ammonium sulfide         (NH₄)₂S

Iron III chloride                FeCl₃

Aluminum Nitrate

Barium acetate               Ba(C₂H₃O₂)₂

Calcium bromide

Sodium nitrate

Lithium oxide               Li₂O

Copper II bromide       CuBr₂

Calcium hydroxide

Ammonium phosphate

Magnesium nitrate

Strontium nitrate          Sr(NO₃)₂

Silver sulfide                 Ag₂S

Cobalt III hydroxide      Co(OH)₃

Copper II chloride

Potassium sulfide

Iron III sulfate

Mg₃(PO₄)₂

Sn(NO₃)₄

Sodium phosphate

Generally, the solubility of solids in water increases with increasing temperature. Nevertheless, there are exceptions with some solids and gases. Temperature can be used to create supersaturated solutions, where more solute is dissolved than the equilibrium solubility.

In terms of the solubility of solids in water, generally the solubility increases with increasing temperature. This can be verified by the general trend observed in the solubility curves of various substances, as represented in Figure 11.16. However, there are exceptions to this typical behavior, such as the ionic compound cerium sulfate, which is less soluble at higher temperatures.

The temperature dependence of solubility can be applied in creating supersaturated solutions, where a higher concentration of solute is retained in the solution than what is achievable at equilibrium. This is done by dissolving the solute at a higher temperature and then cooling the solution without allowing the excess solute to precipitate. In some instances, the solubility of gases in water decreases with increasing temperatures.

The dissolution process is influenced by several factors, including temperature, pressure, and the nature of the solute and solvent. Practically speaking, while some solids are entirely insoluble in water, many others exhibit significant solubility.

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The correct answer is: increases with increasing temperature.

The solubility of the majority of solid substances tends to increase as the temperature increases. This trend can be visualized using solubility curves, which graphically represent the relationship between solubility and temperature.

However, it is essential to note that there are exceptions, as certain compounds may show little change or even a decrease in solubility with rising temperature, such as cerium sulfate.

Examples:

Understanding the general trend and exceptions can help predict the behavior of different substances when dissolved in water at various temperatures.

Answer:

Explanation:

The acidity of a solution is measured by its pH, which is the logarithm of the inverse of the molar concentration of hydronium (H₃O⁺) ions:

When you know the pH value you can find hydronium concentration using the antilogaritm function:

[tex]pH=-log[H_3O^{+}]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-2.50}\\ \\ {[H_3O^+]}=0.0032[/tex]

The unit of molar concentration is M.

To prove your answer you can take the logarithm of 0.0316:

The hydronium ion concentration corresponding to a pH of 2.50 is [tex]{3.16 \times 10^{-3} \text{ M}}\).[/tex]

To find the hydronium ion concentration ([H3O+]), we use the definition of pH, which is the negative logarithm (base 10) of the hydronium ion concentration:

[tex]\[ \text{pH} = -\log_{10} [\text{H}_3\text{O}^+] \][/tex]

Given the pH is 2.50, we can rearrange the equation to solve for [H3O+]:

[tex]\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \] \[ [\text{H}_3\text{O}^+] = 10^{-2.50} \][/tex]

Now, we calculate the value:

[tex]\[ [\text{H}_3\text{O}^+] = 10^{-2.50} = 10^{0.50} \times 10^{-3} \] \[ [\text{H}_3\text{O}^+] = \sqrt{10} \times 10^{-3} \] \[ [\text{H}_3\text{O}^+] \approx 3.16 \times 10^{-3} \text{ M} \][/tex]

Therefore, the hydronium ion concentration, which is also the muriatic acid concentration, is approximately [tex]\(3.16 \times 10^{-3}\)[/tex]M when the pH is 2.50.

Answer:

Double replacement reaction

Explanation:

Now, let us first write the reaction equation properly:

     H₂SO₄ + 2KOH ⇒ K₂SO₄ + 2H₂O

The above reaction is a neutralization reaction between an acid and a base whose product gives salt and water only at most instances.

From here, we can observe that the species displaces on another in their ionic state. Hydrogen replaces potassium and water is produced. Potassium combines chemically with sulfate ions to give the salt of potassium.

Answer:

[tex][OH^-]^2=2.0\times 10^{6}[/tex]

Explanation:

The ionization constant of the water = [tex]4.0\times 10^{12}[/tex] at a temperature of [tex]112\ ^0C[/tex] and pressure of 20 MPa

Water dissociates as;

[tex]H_2O\rightleftharpoons H^++OH^-[/tex]

The expression for the ionization constant is :

[tex]K_w=[H^+][OH^-][/tex]

For pure water,

[tex][H^+]=[OH^-][/tex]

So,

[tex]K_w=[OH^-]^2[/tex]

[tex][OH^-]^2=4.0\times 10^{12}[/tex]

The concentration of the hydroxide ions, [tex][OH^-]^2=2.0\times 10^{6}[/tex].

Explanation:

The given precipitation reaction will be as follows.

        [tex]AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)[/tex]

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

                    Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]

                     0.269 M = [tex]\frac{\text{no. of moles}}{1.14 L}[/tex]

                    no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

               No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                   0.307 mol = [tex]\frac{mass}{143.32 g/mol}[/tex]

                           mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.

Answer:

45.3 °C was the initial temperature

Explanation:

Step 1: explain the problem

We have to find the initial temperature, when a certain amount of heat raises this sample of water to its boiling point ( 100 °C)  

⇒this amount of heat = 5.61 x 10^5 J

We will use the formule : Q = mcΔT

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with ΔT = change in temperature ( in °C or K)

The  specific heat of water is 4.186 J/g °C

Step 2 : Calculate the initial temperature

To find the initial temperature, we have to rearrange the formule:

ΔT = Q / mc

In this case we have :

ΔT = 5.61 * 10^5 J / 2450 g * 4.186 J/g °C = 54.70

⇒ The final temperature of the water is the boiling point (100 °C). The change of temperature is 54.70  This means that the boiling point is 54.70°C higher than the initial temperature.

This means : ΔT = Tboiling point - Tinitial

ΔT = 54.70 °C = 100 °C - Tinitial

Tinitial = 100 °C - 54.70 °C = 45.3 °C

The initial temperature of the water is 45.3 °C

Answer:

The answer to your question is: 53.46 % pure

Explanation:

data

Fe = 1.92 ✕ 103 kg   produced    = 1920 kg

Fe2O3 = 5.13 ✕ 103 kg    sample   = 5130 kg

MW Fe2O3 = (56x2)+(16x3) = 160 kg

% of purity = ?

                   Fe2O3 + 3 CO → 2 Fe + 3 CO2

Convert mass to moles

Fe

      56 kg  --------------------- 1 mol

  1920 kg ---------------------   x moles       x = 34.28 moles

From the reaction

   1 mol of Fe2O3 ---------------------  2 moles of Fe

      x moles of Fe2O3 --------------- 34.28 moles

  x = 34.28/2 = 17.14 moles of Fe2O3

       160 kg of FE2O3 ----------------  1 mol

             x kg of Fe2O3---------------  17.14 moles

x = 17.14 x 160/1 = 2742,4 kg of Fe2O3  It's supposed to be the amount of Fe if it was 100% pure.

                             2742.4 kg of Fe2O3 ----------------  100%

                              5130 kg of Fe2O3 -------------------   x

x = (2742.4x100)/5130 = 53.46 pure

Answer:

b. H3O c. H2SO4 d. HNO3

Explanation:

The hydrogen bonds can be form with molecules that have free electron pairs  and hydrogen atoms. In the case of NH4 molecule, the "free" electron pair in the nitrogen is not free because there is an hydrogen occuping them, and, in fact, this molecule has a positive charge. The other acids have free electron pairs in the oxigen, nitrogen or sulfur.

The molecules that have the ability to form hydrogen bonds with themselves

are H₃0, H₂SO₄, and HNO₃

The above element removes the high number of the electron density in the covalent bond with hydrogen, leaving the H atom lack or lower number of electron

An hydrogen bond is also defined as the electromagnetic attraction that is formed between a partially positively charged hydrogen atom fused to a highly electronegative atom and another close electronegative atom.

Conclusively, we can therefore say that option b, c, and d explain what the statement means.

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Answer:

c. temperature and water vapor content.

Explanation:

The density of any substance is dependent of temperature, and in gases, the variations is greater than in any other substance, therefore, temperature is very important to calculate air density. Also, the concentration of water vapor in the air can vary depending  where is located, and that concentration will affect the density because it affects the average mass of air.

Answer:

c. temperature and water vapor content is the correct answer.

Explanation:

temperature and water vapor content are the two most important in determining the density of air.

When the temperature gets increase it results in the increase of volume and density gets decreases.

Temperature is an important factor to determine the air density.

The concentration of water vapor in the air change depending on where it is placed and so when water vapor is added it will change the density of air because it changes the average mass of air and density gets decrease.

Answer:

The answer to your question is: less than

Explanation:

Some properties of matter depend on the amount of matter in a sample others do not.

Extensive properties: are properties that depend on the amount of matter in a sample. Ex. size, volume

Intensive properties: are properties that do not depend on the amount of matter in a sample. Ex. color, temperature, density, solubility, boiling point.

Then in your question boiling point does not depend on the amount of matter, then the boiling points will be the same as in number 1.

The boiling point of a substance is the temperature at which the substance changes to vapour. The value of M in the scenario given will be less than N

Given equal mass ;

Hence, 100°C < 300°C ; M < N

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Answer : The enthalpy change of reaction is -23.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)[/tex]    [tex]\Delta H=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)[/tex]    [tex]\Delta H^o_1=-824.2kJ[/tex]

(2) [tex]CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)[/tex]    [tex]\Delta H^o_2=-282.7kJ[/tex]

First we will reverse the reaction 1 and multiply equation 2 by 3 then adding both the equation, we get :

(1) [tex]Fe_2O_3(s)\rightarrow 2Fe(s)+\frac{3}{2}O_2(g)[/tex]    [tex]\Delta H^o_1=+824.2kJ[/tex]

(2) [tex]3CO(g)+\frac{3}{2}O_2(g)\rightarrow 3CO_2(g)[/tex]    [tex]\Delta H^o_2=3\times (-282.7kJ)=-848.1kJ[/tex]

The expression for final enthalpy is,

[tex]\Delta H=\Delta H^o_1+\Delta H^o_2[/tex]

[tex]\Delta H=(+824.2kJ)+(-848.1kJ)[/tex]

[tex]\Delta H=-23.9kJ[/tex]

Therefore, the enthalpy change of reaction is -23.9 kJ

Acidity.  As you go down it’s more of an acid as you go up it becomes more of a base

Final answer:

A decrease in pH from 9 to 8 indicates an increase in the concentration of [OH-] tenfold (10X).

Explanation:

The pH of a solution is a measure of how acidic or basic it is. A pH of 9 indicates a basic solution, while a pH of 8 indicates a slightly less basic solution. As the pH decreases from 9 to 8, the concentration of [OH-] increases tenfold (10X) compared to what it was at pH 9.

Answer:

47 g/mol

Explanation:

0.70 (50 amu) + 0.25 (35 amu) + 0.05 (70 amu) = 47 amu = 47 g/mol

Explanation:

The atomic weight of element X is calculated by taking the weighted average of the atomic masses of its isotopes. Considering the individual contribution of each isotope, which is determined by multiplying each isotope's atomic mass by its relative abundance, the atomic weight of element X is 47.25 amu.

The atomic weight of an element is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes. An isotope's contribution to the overall atomic weight is proportional to its abundance. In this case, Element X is composed of three isotopes A, B, and C.

Following are the contributions from each isotope:

So, by adding all these contributions, the atomic weight of element X would be 35 + 8.75 + 3.5 = 47.25 amu.

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Final answer:

To determine the grams of CO₂ produced in the reaction between C₆H₆ and O₂, we need to balance the chemical equation and use mole ratios.

Explanation:

To determine how much CO₂ will be produced in the reaction between C₆H₆ and O₂, we first need to balance the chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

The balanced equation tells us that 2 moles of C₆H₆ react with 15 moles of O₂ to produce 12 moles of CO₂. To find the amount of CO₂ produced, we can use the given masses of C₆H₆ and O₂ to calculate their respective moles:

42.5 g C₆H₆ x (1 mol C₆H₆ / 78.114 g C₆H₆) = 0.544 mol C₆H₆

113.1 g O₂ x (1 mol O₂ / 32.00 g O₂) = 3.538 mol O₂

Since the mole ratio of C₆H₆ to CO₂ is 2:12, we can set up a proportion to find the moles of CO₂:

0.544 mol C₆H₆ x (12 mol CO₂ / 2 mol C₆H₆) = 3.264 mol CO₂

Finally, we can convert the moles of CO₂ to grams using its molar mass:

3.264 mol CO₂ x (44.009 g CO₂ / 1 mol CO₂) = 143.58 g CO₂

Therefore, 143.58 grams of CO₂ will be produced in this reaction.

The amount of CO2 produced by this reaction is 124.44 grams.

To solve this problem, we need to follow these steps:

1. Write down the balanced chemical equation for the combustion of benzene (C6H6).

2. Calculate the moles of C6H6 and O2 using their molar masses.

3. Determine the limiting reactant, which is the reactant that will be completely consumed first and thus limit the amount of product formed.

4. Use the stoichiometry of the balanced equation to calculate the mass of CO2 produced from the limiting reactant.

The balanced chemical equation for the combustion of benzene is:

[tex]\[ C_6H_6 + \frac{15}{2}O_2 \rightarrow 6CO_2 + 3H_2O \][/tex]

Calculate the moles of C6H6 and O2:

[tex]The molar mass of C6H6 is \( 6 \times 12.01 \, g/mol \, (C) + 6 \times 1.008 \, g/mol \, (H) = 78.11 \, g/mol \).[/tex]

The moles of C6H6 are:

[tex]\[ \frac{42.5 \, g}{78.11 \, g/mol} \approx 0.544 \, mol \][/tex]

The molar mass of O2 is [tex]\( 2 \times 16.00 \, g/mol = 32.00 \, g/mol \).[/tex]

The moles of O2 are:

[tex]\[ \frac{113.1 \, g}{32.00 \, g/mol} \approx 3.534 \, mol \][/tex]

Determine the limiting reactant:

From the balanced equation, the stoichiometric ratio of C6H6 to O2 is 1:15/2, which simplifies to 2:15 or 4:30. To react with 0.544 mol of C6H6, we would need:

[tex]\[ 0.544 \, mol \, C_6H_6 \times \frac{30}{4} = 4.08 \, mol \, O_2 \][/tex]

Since we only have 3.534 mol of O2, O2 is the limiting reactant.

Calculate the mass of CO2 produced:

From the balanced equation, the stoichiometric ratio of O2 to CO2 is 15/2:6. Therefore, the moles of CO2 produced are:

The molar mass of CO2 is [tex]\( 12.01 \, g/mol \, (C) + 2 \times 16.00 \, g/mol \, (O) = 44.01 \, g/mol \).[/tex]

The mass of CO2 produced is:

[tex]\[ 2.827 \, mol \, CO_2 \times 44.01 \, g/mol = 124.44 \, g \][/tex]

Answer:

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu)  ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂)  / 100

%₁ = x I established this is an equation

%₂ = 100 - x

Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100    and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x  /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

c)95.7%

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu)  ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula to be used to calculate the percentage for isotopes will be:

Atomic mass = m₁(%₁) +m₂(%₂)  / 100

%₁ = x

%₂ = 100 - x

On substituting the values:

[tex]114.82 = \frac{112.8040x + 114.9039(100-x)}{100} \\\\ 114.82 = \frac{ 112.8040x + 11490.39 - 114.9039x }{100} \\\\ 114.82 = 112.8040x -114.9039x +11490.39\\\\ 114.82 - 11490.39 = 112.8040x -114.9039x\\\\ -8.39 = -2.099x\\\\ x = 3.99 [/tex]

Thus, % of Indium-115 = 100 - 3.99 = 96%.

Therefore, the correct option is c.

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Answer:

The answer to your question is Endothermic reaction

Explanation:

Endothermic reactions are reactions that absorbe energy from the environment.

All the reactions that needs heat to proceed are endothermic, example: photosynthesis, reactions in the laboratory that are heated, heat our meals.

This is an endothermic reaction because the enthalpy of the products is higher than the enthalpy of the reactants,

A reaction diagram illustrates the changes in enthalpy and the reaction progression, with the 'hill' or 'saddle' shapes indicating the activation energy barriers in one-step or two-step reactions, respectively.

A reaction diagram illustrates the changes in enthalpy and the progression of a chemical reaction over time. The diagram you're describing, with either a 'hill' shape or a 'saddle' shape, represents the energy changes that occur during a chemical reaction. The vertical axis measures the enthalpy of the system, while the horizontal axis represents the reaction coordinate, showing the progress from reactants to products.

In a one-step reaction, the diagram shows one energy barrier to be overcome, which is represented by the 'hill' on the curve. This barrier is known as the activation energy, which is the energy required to initiate the reaction. In a two-step reaction, two such barriers exist, and the 'saddle' shape of the reaction curve reflects this. The highest points on these curves represent the transition states, which are unstable species that exist at the peak of the activation energy.

The type of reaction can often be determined by the relative energies of the reactants and products. If the products have lower energy than the reactants, the reaction is exothermic. Conversely, if the products have higher energy, the reaction is endothermic. The difference in enthalpy between the reactants and products indicates the overall energy change of the reaction.

Answer: The correct answer is Option 5.

Explanation:

[tex]M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of the NaOH.

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of the [tex]Ca(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL[/tex]  

Putting all the values in above equation, we get:

[tex]M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M[/tex]

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base.

We are given:

[tex]n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M[/tex]

Hence, the correct answer is Option 5.

The molarity of the HCl solution used to neutralize a basic solution of NaOH and Ca(OH)2 is 0.060 M. The correct answer is option 5.

The question asks about the molarity of the HCl solution used to neutralize a basic solution of NaOH and Ca(OH)2. The first step is to determine the moles of OH- ions contributed by both NaOH and Ca(OH)2. Next, the moles of HCl need to be found because in a neutralization reaction, the moles of H+ ions (from HCl) will equal the moles of OH- ions (from NaOH and Ca(OH)2). The molarity of the HCl is then calculated by dividing the moles of HCl by the volume of the HCl solution which is given in mL but needs to be converted to L for this calculation.

First, calculate the number of moles from each OH- source: NaOH (0.10 M x 0.050 L = 0.005 moles) and Ca(OH)2 (2 x (0.10 M x 0.050 L) = 0.01 moles) because each formula unit of Ca(OH)2 supplies 2 OH- ions. So, total moles of OH- = 0.005 moles + 0.010 moles = 0.015 moles.

Since the moles of H+ ions (from HCl) equals the moles of OH- ions (from NaOH and Ca(OH)2), the moles of HCl also equals 0.015 moles. Finally, calculate the molarity of the HCl solution: Molarity = Moles / Volume = 0.015 moles / 0.250 L = 0.060 M.

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Answer:

ΔG° = –7.4 kcal/mole

Explanation:

When there are coupled reactions taking place, we can say that their free energy variations are additive. Meaning that if there is a reaction A → B followed by B → C, each reaction will have its own free energy variation, ΔG°₁ and ΔG°₂. As both reactions are secuential, there is a global reaction taking place, which is A → C. This global reaction also has its own free energy variation, ΔG°total, which represents the sum of the individual free energy variations of the coupled reactions, ΔG°₁ +  ΔG°₂ = ΔG°total.

For this case we have, in the first reaction of glycolysis, there are two coupled reactions that take place:

(1) glucose + Pi → glucose 6-phosphate + H₂O

(2) ATP + H₂O → ADP + Pi

We know that the total variation of free energy for that reaction is:

ΔG°total =  –4.06 kcal/mole

But the individual variations of free energy, ΔG°₁ and ΔG°₂, are unknown, so we can propose the next equation:

ΔG°₁ + ΔG°₂ = –4.06 kcal/mole

Then we have another value given to us, the variation of free energy of the reversed first reaction (1), which would be:

(3) glucose 6-phosphate → glucose + Pi

ΔG°₃ = –3.34 kcal/mole

As this reaction is the reversed reaction of the first one (1), we can assume the next:

ΔG°₁  = (–1) * (ΔG°₃) = (–1) * (–3.34 kcal/mole) = 3.34 kcal/mole

So now that we have the value of ΔG°₁ we can substitute it in the proposed equation to find out the value of ΔG°₂ :

ΔG°₁ + ΔG°₂ = –4.06 kcal/mole

ΔG°₂ = –4.06 kcal/mole – ΔG°₁

ΔG°₂ = –4.06 kcal/mole – 3.34 kcal/mol

ΔG°₂ = –7.4 kcal/mole

So there it is, that is the value of the variation of free energy of the second reaction (2), which is the hydrolisis of ATP.

Final answer:

The ΔG° for hydrolysis of ATP to ADP and Pi, considering the ΔG° values given for the phosphorylation and dephosphorylation of glucose, is -7.40 kcal/mole.

Explanation:

The student asked for the ΔG° for the hydrolysis of ATP. In enzymatic reactions, the ΔG° of the reverse reaction is the negative of the ΔG° of the forward reaction. Given that hexokinase catalyzes the phosphorylation of glucose to glucose 6-phosphate with a ΔG° of -4.06 kcal/mole and the reverse reaction, catalyzed by glucose 6-phosphatase, has a ΔG° of -3.34 kcal/mole, we can find the ΔG° for hydrolysis of ATP by adding these two values (because to find the overall ΔG° for a coupled reaction, you sum the ΔG° values of the individual steps). The sum of -4.06 kcal/mole and -3.34 kcal/mole is -7.40 kcal/mole, which is the ΔG° for the hydrolysis of ATP to ADP and Pi.

Answer:

189.91g

Explanation:

Given parameters:

Mass of beaker with oil = 200.6g

Mass of beaker = 10.69g

Unknown:

Mass of corn oil = ?

Solution:

To find the mass of corn oil, we know that:

 Mass of beaker with oil = Mass of beaker + mass of corn oil

Therefore:

           Mass of corn oil = Mass of beaker with oil - Mass of beaker

    Mass of corn oil = (200.60 - 10.69)g

    Mass of corn oil = 189.91g

Answer:

From hypotonic to hypertonic

Explanation:

Water diffusion is a phenomenon that occurs when a solute (eg. a salt) is present in different concentrations in different areas. Because the concentration is inversely proportional to volume (meaning that the higher the volume, the lower the concentration), water will move from areas with lower concentration (hypotonic) to areas with higher concentration (hypertonic), so as to match the concentrations.

Answer:

Water will move from hypotonic to hypertonic solutions

Explanation:

When a hypotonic solution is separated by a permeable membrane from another solution with more solute (hypertonic), water will move across the membrane until both solutions have the same concentration: when water comes into the compartment with hypertonic solution, this solute dissolves.

Water will not move from hypertonic to hypotonic without energy addition.

Answer:

D. All of these

Explanation:

Hi,

So, let's go,

The first thing to know, is that strong acids has a lot of H+ free in solution. While weak acids doesn't have a lot of H+ free.

Now, analyzing the letters

A. Electrical current conduction has a lot of free electrons, so, strong acids has a lot of H+ free in solution too. The combination of these two, in large amount, results in a strong reaction. Very differently with weak acids, which has a few free electrons.

B. ph paper test works with a scale of colors. This means that at one end the acid is strong, and at the other the acid is weak. So, if you drip a drop of the strong acid, the ph paper test will show you different color that when you drop a weak acid.

c. Magnesium is an element from IIA family, that means he reacts strongly with electrons and íons. Which is the cafe os strong acids.

I hope that all questions is solved.

Bye! See ya

Answer:

B.  1.7 g/cm^3.

Explanation:

Density = mass / volume

= 520 / 300

=  1.733.

The density of a substance is calculated by dividing mass by the volume of the object. A substance weighing 520 grams has a density of 1.7 g/cm³. Thus, option B is correct.

Density has been constituted of mass in grams per cubic centimeter of substance. The volume inversely affects the density possessed by the substance. It gives the estimation of buoyancy and whether the object will float or sink.

It is calculated with the help of mass and volume and its formula is given as,

Density (D) = Mass (m) ÷ Volume (V)

Given,

Volume (V) = 300 cubic centimeter

Mass (m) = 520 grams

Density is calculated as,

Density (ρ) = Mass (M) ÷ Volume (V)

ρ = 520 ÷ 300

ρ =  1.733 grams per cubic centimeter

Therefore, option B. 1.7 grams per cubic centimeter is the density of the substance.

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To determine the limiting reactant, compare the moles of N2 and H2. The excess reactant can be determined by subtracting the moles of the limiting reactant used from the moles of the excess reactant initially present. The amount of NH3 formed can be calculated using the stoichiometry of the balanced equation.

To determine the limiting reactant in this reaction, we need to compare the moles of N2 and H2. The balanced equation tells us that it takes 1 mole of N2 to react with 3 moles of H2 to produce 2 moles of NH3. Therefore, if we have 2.00 moles of N2 and 4.00 moles of H2, the limiting reactant is the one that runs out first. Since 1 mole of N2 reacts with 3 moles of H2, we need 6.00 moles of H2 to react completely with 2.00 moles of N2. The excess reactant can be determined by subtracting the moles of the limiting reactant used from the moles of the excess reactant initially present. In this case, the excess reactant is H2, and the amount of excess H2 remaining can be calculated. Lastly, to determine the amount of NH3 formed, we use the stoichiometry of the balanced equation. Since the balanced equation tells us that 1 mole of N2 produces 2 moles of NH3, we can calculate the moles of NH3 formed from the moles of N2 used.

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Answer: The osmotic pressure is 54307.94 Torr.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iCRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = ?

i = Van't hoff factor = 3

C = concentration of solute = 0.958 M

R = Gas constant = [tex]62.364\text{ L Torr }mol^{-1}K^{-1}[/tex]

T = temperature of the solution = [tex]30^oC=[30+273]K=303K[/tex]

Putting values in above equation, we get:

[tex]\pi=3\times 0.958mol/L\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 303K\\\\\pi=54307.94Torr[/tex]

Hence, the osmotic pressure is 54307.94 Torr.

Final answer:

The osmotic pressure of the 0.958 M solution is 23.96 atm.

Explanation:

The osmotic pressure of a solution can be calculated using the formula II = MRT, where II is the osmotic pressure, M is the molarity of the solution, R is the gas constant (0.08206 L atm/mol K), and T is the temperature in Kelvin. In this case, we have a 0.958 M solution with a volume of 6.00 L at a temperature of 30 °C. To convert the temperature to Kelvin, we add 273 to get 303 K. Substituting the values into the formula, we get:

II = (0.958 M) (0.08206 L atm/mol K) (303 K)

II = 23.96 atm

Therefore, the osmotic pressure of the solution is 23.96 atm.

The new solution with 40 grams of solute in 160 grams of solvent is saturated, as it has the same concentration of solute per 100 grams of solvent as the original saturated solution.

When a solution reaches the point where no additional solute will dissolve, it is considered saturated. In the student's original solution, 25 grams of solute in 100 grams of solvent represents a saturated solution. If a new solution contains 40.0 grams of solute in 160 grams of solvent, we must compare the concentration of solute to the solvent's ability to dissolve that solute at a given temperature. Assuming temperature is constant and solubility does not change, the new solution can be compared to the original by calculating the concentration of solute per 100 grams of solvent. In the original solution, the saturated concentration is 25 grams of solute per 100 grams of solvent. In the new solution, adjusting to per 100 grams of solvent, we have 25 grams of solute per 100 grams of solvent (since (40 grams/160 grams)*100 equals 25). This means the new solution has the same concentration as the original saturated solution, and thus, it is also saturated.

Answer:

The respective figure with label and targets is missing but yet the definitions and stability considerations can help you, so I explain them below.

Explanation:

Remember these definitions:

Hence,

The neutrons provide stability to the nucleus of the atom by compensating the electrostatic repulsion force that arise from the fact that positive charges are forced to be so close in the nucleus.

Since the more protons are added to the nucleus the stronger the repulsive force inside the nucleus are, as the atomic number increase the neutron number must increase too.

For the ligther elements (lower atomic and mass numbers) the ratio of neutrons to protons is very close to 1.

For heavier elements (greater atomic and mass numbers) the ratio of neutrons to protons increase: proportionally more neutrons are needed to provide stability to the nucleus.

Final answer:

The atomic number is defined by the number of protons, and an element's mass number is the total of its protons and neutrons. Isotopes have different numbers of neutrons, thus changing the mass number, while the atomic number remains constant. The atomic mass is an average of the mass numbers of an element's isotopes.

Explanation:

The atomic number and mass number of an element are essential concepts in chemistry. The atomic number of an element is defined by the number of protons in its nucleus and is used to distinguish one element from another. The mass number, on the other hand, is the sum of the number of protons and neutrons within the nucleus. It is important to note that electrons contribute negligibly to an element's mass and are, therefore, often disregarded when calculating the mass number.

Isotopes are different forms of the same element that vary only in their number of neutrons. This variability can change the mass number, and thus, isotopes of the same element will have slightly different mass numbers. Since the atomic number remains unchanged (protons do not change), the number of neutrons can be calculated by subtracting the atomic number from the mass number. The atomic mass of an element is a weighted average that reflects the mass numbers of all its naturally occurring isotopes, and is often a decimal number as a result.