Write the balanced chemical equation for the combustion of ethane, , and answer these questions. (Use the lowest possible coefficients. Omit states of matter.) How many molecules of oxygen would combine with 16 molecules of ethane in this reaction

Answer:

ion travelled : 2.00cm -0.40cm

                     = 1.6cm

∴ Rf = 1.6/2.0

        = 0.80

Explanation:

Retention factor is the ratio of distance travelled by solute divided by distance travelled by solvent.

since you given distance travelled by solvent then the distance required by solute is need in this case the ion.

Answer:

The procedure you will use in this exercise exploits the difference in acidity and solubility just described.

(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.

(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.

(c) you will extract with sodium hydroxide to remove any phenol that is present.

(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.

Sulfonation of benzene involves the replacement of a hydrogen atom by a sulfonyl group using sulfuric acid. The reaction involves several steps, each with specific reactants and products. Key substances involved include sulfuric acid, sulfurous acid, benzene, and sulfur dioxide.

The question is asking for a detailed understanding of the sulfonation of benzene, a process where a hydrogen atom on benzene is replaced by a sulfonyl group. This reaction is a type of electrophilic aromatic substitution. Sulfonation involves the use of sulfuric acid, which provides H3O+ and SO3 in the first reversible step. In the second step, SO3 reacts slowly with benzene to form H(C6H5+)SO3−, and in the subsequent fast step, HSO4− reacts with H(C6H5+)SO3− to yield C6H5SO3− and H2SO4. Finally, C6H5SO3− and H3O+ react to produce the sulfonated benzene product, C6H5SO3H.

The reaction mechanism involves various substances, including sulfurous acid and sulfur dioxide. Sulfurous acid is unstable and can decompose into sulfur dioxide and water. Sulfur dioxide can further react with oxygen to form sulfur trioxide, which can participate in the sulfonation of benzene. Note that sulfur forms several compounds exhibiting positive oxidation states, unlike oxygen.

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Answer:

Ar < Cl - < S2-

Explanation:

All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.

However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.

The correct order of increasing expected size among Cl-, S2-, and Ar based on the simple shell model and their respective nuclear charges—considering they are all isoelectronic—is Ar < Cl- < S2-.

The question asks to arrange the atoms and/or ions Cl-, S2-, and Ar in order of increasing expected size based on the simple shell model of atomic radius. In the simple shell model, atomic radius increases with the addition of electron shells and decreases with an increase in nuclear charge for isoelectronic species, meaning species with the same number of electrons. Cl-, S2-, and Ar are all isoelectronic with the closed-shell electron configuration of [Ar]. The Cl- ion has 17 protons, S2- has 16 protons, and Ar has 18 protons.

Following this reasoning:

Therefore, the correct order from smallest to largest is: Ar < Cl- < S2-.

Answer :  The rate of reaction is,

[tex]Rate=4.77\times 10^{-19}M/s[/tex]

The appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]

Explanation :

The general rate of reaction is,

[tex]aA+bB\rightarrow cC+dD[/tex]

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

[tex]\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}[/tex]

[tex]\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]

[tex]\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}[/tex]

[tex]\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]

[tex]Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

[tex]NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)[/tex]

The rate law expression will be:

[tex]Rate=k[NO_2][O_3][/tex]

Given:

Rate constant = [tex]k=1.69\times 10^{-4}M^{-1}s^{-1}[/tex]

[tex][NO_2][/tex] = [tex]1.77\times 10^{-8}M[/tex]

[tex][O_3][/tex] = [tex]1.59\times 10^{-7}M[/tex]

[tex]Rate=k[NO_2][O_3][/tex]

[tex]Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})[/tex]

[tex]Rate=4.77\times 10^{-19}M/s[/tex]

The expression for rate of appearance of [tex]NO_3[/tex] :

[tex]\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}[/tex]

As, [tex]\text{Rate of reaction}=4.77\times 10^{-19}M/s[/tex]

So, [tex]\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s[/tex]

Thus, the appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]

The rate of reaction is [tex]4.77*10^{-19} M/s[/tex]

The rate of the appearance of [tex]NO_{3}[/tex] is, [tex]4.77*10^{-19} M/s[/tex]

The second order reaction is given as,

                     [tex]NO_{2}(g) + O_{3}(g) \Rightarrow NO_{3} (g) + O_{2} (g)[/tex]

Given that,

      Rate constant [tex]k=1.69*10^{-4}[/tex]

       [tex][NO_{2}] = 1.77 *10^{-8} M\\\\O_{3}= 1.59 * 10^{-7} M[/tex]

Rate of reaction is given as,

             [tex]Rate=k[NO_{2}][O_{3}]\\\\Rate=1.69*10^{-4}* 1.77*10^{-8}*1.59*10^{-7}\\\\Rate=4.77*10^{-19} M/s[/tex]

the rate of the appearance of [tex]NO_{3}[/tex] is,

              [tex]\frac{d[NO_{3}]}{dt}=4.77*10^{-19} M/s[/tex]

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Answer:

The answer is 3-Phenylpropanoic acid (see attached structure)

Explanation:

From spectral data:

3005 cm-1 ⇒ carboxylic acid (broad band)

1670 cm-1 ⇒ C=C

1603 cm-1 ⇒ Aromatic C-C bond

H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding  (clearer investigation of  spectrum would be expedient).

Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.

The compound is probably a molecule with an aromatic ring (benzene) with an attached ester group (COOCH3) that satisfies all the given spectral data.

The spectral data given corresponds to a compound with molecular formula C9H10O2. From the IR data, the bands at 3005 cm-1 suggests C-H sp2 bond (alkene or aromatic), at 1676 cm-1 indicates carbonyl group (C=O), and 1603 cm-1 suggests a carbon-carbon double bond (C=C) which might be in an aromatic ring. The 1H NMR data: 2.6 ppm singlet signifies the protons of a methyl group (CH3) attached to an electronegative atom like a carbonyl carbon.

Based on these data, a probable structure for the compound could be a molecule, which is an aromatic ring (benzene) with an attached ester group (COOCH3). That gives the right molecular formula, the required carbonyl, alkene and sp2 hybridized C-H bonds for the IR, and the singlet in the NMR for the methyl group (CH3) of the ester.

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Final answer:

The mass 0076.21 g should be rounded to two significant figures as 76 g, omitting leading zeros, which are not significant. Significant figures indicate precision and uncertainty in scientific measurements.

Explanation:

The digital scale used by the chemist displays the mass of the sample as 0076.21 grams. To report the mass with two significant digits, we round the figure to 76 grams. This is because the leading zeros do not count towards significant figures; they are only placeholders to indicate the decimal position. The '7' and '6' are the two significant digits from the number 0076.21 grams.

It is crucial to use the correct number of significant figures in scientific measurements because they communicate the precision of the measurement and the uncertainty. For instance, if a balance measures to the nearest ±0.1 mg, one should not report a mass with more precision than this, as it would suggest a false level of accuracy. For addition and subtraction, the result must not have more decimal places than the measurement with the least precision. So if the scales used have different precisions, the final weight should be reported with the precision of the least precise scale.

The covalent character of a bond usually increases with the electronegativity difference between the atoms involved in the bond. Bonds between nonmetals usually have high covalent character, whereas bonds between a metal and a nonmetal are often ionic. Therefore, the bond pairs can be ordered in increasing covalent character as: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F).

The covalent character of a bond generally increases with the electronegativity difference between the two atoms involved in the bond. Using this principle, we can establish that the bonds between H and F, Si and Cl, C and H, and H and I have more covalent character compared to the rest because the pairs consist of nonmetals. The bond between K and Br is likely to have the most ionic character, as K (potassium) is a metal, and Br (bromine) is a nonmetal. This is due to the general guideline that bonds between a metal and a nonmetal are often ionic.

However, it's also important to understand that while this rule can be generally applied, there are many exceptions and other factors can influence the character of a bond.

So considering this, the bonds in increasing order of covalent character might be: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F)

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In the given list of atoms, the bonds they form increase in covalent character as follows: (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is based on factors including the atoms' electronegativity differences and bond orders.

The subject here is the covalent character of different bonds, which can be decided based on multiple factors such as electronegativity differences and the bond order.

Electronegativity is the measure of the tendency of an atom to attract shared electrons. Bonds between atoms with a large eleber of chemical bonds between a pair of atoms - greater bond orders typically result in stronger, more covalent bonds. Given these criteria, the order of increasing covalent character among the given pairs should be (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is because K and Br have a large electronegativity difference making the bond more ionic, while Se and S are both non-metals with similar electronegativities, which makes their bond more covalent.ctronegativity difference are usually more ionic and less covalent. On the other hand, bond order refers to the num

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Answer:

pOH = 4.8

pH = 9.2

Explanation:

Given data:

Hydrogen ion concentration = 6.3×10⁻¹⁰M

pH of solution = ?

pOH of solution = ?

Solution:

Formula:

pH = -log [H⁺]

[H⁺] = Hydrogen ion concentration

We will put the values in formula to calculate the pH.

pH = -log [6.3×10⁻¹⁰]

pH = 9.2

To calculate the pOH:

pH + pOH = 14

We will rearrange this equation.

pOH = 14 - pH

now we will put the values of pH.

pOH = 14 - 9.2

pOH = 4.8

Answer:

47.47 g of CO₂ is the amount formed.

Explanation:

The reaction is:

2C₆H₅COOH(aq) + 15O₂(g) → 14CO₂(g) + 6H₂O(l)

Let's apply the formula for the percent yield

Percent yield of reaction = (Produced yield/Theoretical yield) . 100

First of all we convert the moles of CO₂ to mass: 1.30 mol . 44 g /1 mol = 57.2 g. So now, we replace:

(Produced yield / 57.2 g ). 100 = 83

Produced yield / 57.2 g  = 83 / 100

Produced yield / 57.2 g  = 0.83

Produced yield = 0.83 . 57.2g → 47.47 g of CO₂

Answer:

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Explanation:

Step 1: Data given

Number of CO2 = 1.30 moles

Percent yield = 83.0 %

Step 2: The balanced equation

15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)

Step 3: Calculate the number moles of CO2 formed

1.30 moles CO2 = 100 %

The actual amount of moles = 0.83 * 1.30 = 1.079 moles

Step 4: Calculate the percent yield of the reaction

We can control this by calculating the percent yield of the reaction

% yield = (actual yield / theoretical yield ) * 100 %

% yield = (1.079 moles / 1.30 moles ) * 100 %

% yield = 83.0 %

Step 5: Calculate thr mass of CO2 produced

Mass CO2 = moles * molar mass CO2

Mass CO2 = 1.079 moles * 44.01 g/mol

Mass CO2 = 47.5 grams

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Answer:

The equilibrium partial pressure of NO2 is 0.152 atm

Explanation:

Step 1: Data given

Initial pressure of NO2 = 0.500 Atm

Total pressure inside the vessel at equilibrium = 0.674 atm

Step 2: The balanced equation

2 NO2(g) ⇌ 2 NO(g) + O2(g)

Step 3: The initial pressures

pNO2 = 0.500 atm

pNO = 0 atm

pO2 = 0 atm

Step 4: The pressure at the equilibrium

pNO2 = 0.500 - 2x

pNO = 2x

pO2 = x

Total pressure = 0.674 = (0.500 - 2x) + 2x + x

0.674 = 0.500 + x

x = 0.174

pNO2 = 0.500 - 2*0.174 = 0.152 atm

pNO = 2x = 0.348 atm

pO2 = x = 0.176 atm

The equilibrium partial pressure of NO2 is 0.152 atm

The equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂, with a total equilibrium pressure of 0.674 atm and an initial NO₂ pressure of 0.500 atm, is calculated to be 0.152 atm using an ICE table and the stoichiometry of the reaction.

The question involves finding the equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂. Given that the total pressure at equilibrium is 0.674 atm and the initial pressure of NO₂ was 0.500 atm, one can solve for the equilibrium partial pressures using the stoichiometry of the balanced equation and the properties of an equilibrium state.

To determine the equilibrium partial pressure of NO₂, one would set up an ICE (Initial, Change, Equilibrium) table. Denoting the change in NO₂ pressure as x, the change for NO would also be x, and the change for O₂ would be x/2 due to the stoichiometry of the reaction. Given the total pressure at equilibrium, we can solve for x and subsequently find the equilibrium partial pressure of each gas.

Knowing the total pressure and using the ICE table, we can calculate the equilibrium partial pressure of NO₂:
Total pressure at equilibrium = Initial pressure of NO₂ - Change in NO₂ pressure + Change in NO pressure + Change in O₂ pressure

0.674 atm = 0.500 atm - x + x + x/2
0.674 atm = 0.500 atm + x/2
x = 0.348 atm
Equilibrium partial pressure of NO₂ = Initial pressure of NO₂ - Change in NO₂ pressure

0.500 atm - x = 0.500 atm - 0.348 atm = 0.152 atm

To calculate the pH of a buffer solution composed of H2PO−4(aq) and HPO2−4(aq), the specific pKa value for the ionization step of interest needs to be known.

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution.

In the case of phosphoric acid, which is a triprotic acid, we need to use the pKa value that corresponds to the ionization step we are interested in.

Since the question does not specify which buffer composition is being used, we cannot determine which pKa value to use from the information given.

Explanation:

Relation between heat energy and specific heat is as follows.

            [tex]q_{bomb} = c \times \Delta t[/tex]

or,            c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]

where,     c = specific heat

           [tex]q_{bomb}[/tex] = heat energy

           [tex]\Delta t[/tex] = change in temperature

Putting the given values into the above formula we will calculate the specific heat as follows.

                c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]

                   = [tex]\frac{5.23 kJ}{7.33^{o}C}[/tex]

                   = 0.713 [tex]kJ^{o}C[/tex]

Thus, we can conclude that heat capacity C of the calorimeter is 0.713 [tex]kJ^{o}C[/tex].

Final answer:

The heat capacity C of the calorimeter is calculated using the formula C = q / ΔT, with the given values resulting in a C of 0.7138 kJ/°C.

Explanation:

The heat capacity C of the calorimeter can be calculated using the amount of heat energy released and the resultant temperature change. The formula to calculate heat capacity is C = q / ΔT, where q is the heat energy released and ΔT is the change in temperature.

In this case, 5.23 kJ of heat energy was released, causing a temperature rise of 7.33 °C. Therefore, the heat capacity of the calorimeter is calculated as follows:

C = 5.23 kJ / 7.33 °C = 0.7138 kJ/°C

It's crucial to note that the heat capacity is commonly expressed in kJ/°C for bomb calorimeters, reflecting the amount of energy required to raise the temperature of the entire calorimeter setup by one degree Celsius.

Answer:

A. 0.064mol

B. 0.85mol

C. 1500mL

Explanation:

A. Molarity = 0.33M

Volume = 195mL = 195/1000 = 0.195L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.33 x 0.195

Mole = 0.064mol

B. Molarity = 1.7M

Volume = 500mL = 500/1000 = 0.5L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 1.7 x 0.5

Mole = 0.85mol

C. C1 = 12M

V1 = 50mL

C2 = 0.4M

V2 =?

Using the dilution formula C1V1 = C2V2, we find the volume of the diluted solution as follows:

C1V1 = C2V2

12 x 50 = 0.4 x V2

Divide both side by 0.4

V2 = (12 x 50) /0.4

V2 = 1500mL

Answer :

(1) The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(1) The given compound is, [tex]HPO_3^{2-}[/tex]

Let the oxidation state of 'P' be, 'x'

[tex](+1)+x+3(-2)=-2\\\\x=+3[/tex]

The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The given compound is, [tex]Al_2O_3[/tex]

Let the oxidation state of 'Al' be, 'x'

[tex]2x+3(-2)=0\\\\x=+3[/tex]

The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The given compound is, [tex]HIO_4[/tex]

Let the oxidation state of 'I' be, 'x'

[tex](+1)+x+4(-2)=0\\\\x=+7[/tex]

The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

The rules of determining oxidation numbers have been applied to three compounds: hydrogen phosphate (with oxidation numbers of +1 for H, +5 for P and -2 for O), aluminum oxide (+3 for Al and -2 for O), and periodic acid (+1 for H,+7 for I, and -2 for O).

The rules to assign oxidation numbers to the elements in the given compounds may be understood as follows:

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Answer:

a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) 2.40 moles HF

c) 5.25 grams NaF

d)0.609 grams Na2SiO3

e) 0.89 grams Na2SiO3

Explanation:

Step 1: Data given

Step 2: The balanced equation

Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

Mass NaF = 0.125 moles * 41.99 g/mol

Mass NaF = 5.25 grams NaF

d. How many grams of Na2SiO3 can react with 0.800 g of HF?

Moles HF = 0.800 grams / 20.01 g/mol

Moles HF = 0.0399 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol  

Mass Na2SiO3 = 0.609 grams Na2SiO3

Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

Number of moles HF = 0.0399 moles

Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

There will remain 0.0123 - 0.00499 = 0.00731 moles  Na2SiO3

Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

Final answer:

The balanced chemical equation requires 6 moles of HF for every mole of Na2SiO3. For different amounts of reactants, stoichiometry is used to calculate the moles and grams of other substances involved in the reaction.

Explanation:

The balanced chemical equation for the reaction between sodium metasilicate (Na2SiO3) and hydrofluoric acid (HF) is as follows:

Na2SiO3(s) + 6HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O(l)

b. According to the balanced equation, 6 moles of HF are needed to react with 1 mole of Na2SiO3. Therefore, to react with 0.300 mol of Na2SiO3, you would need 1.8 moles of HF (0.300 mol x 6).

c. Again according to the balanced equation, every 6 moles of HF produce 2 moles of NaF. So when 0.500 mol of HF reacts, it produces (0.500 mol HF x 2 mol NaF / 6 mol HF) = 0.167 moles of NaF. The molar mass of NaF is approximately 41.99 g/mol, so the mass of NaF formed is (0.167 mol x 41.99 g/mol) = 7.01 grams.

d. Given the molar mass of HF is approximately 20.01 g/mol, 0.800 g of HF corresponds to (0.800 g / 20.01 g/mol) = 0.0400 moles of HF. Using the balanced equation, we can determine the amount of Na2SiO3 that can react: (0.0400 mol HF x 1 mol Na2SiO3 / 6 mol HF) = 0.00667 moles of Na2SiO3. The molar mass of Na2SiO3 is approximately 122.06 g/mol, so the mass of Na2SiO3 that can react is (0.00667 mol x 122.06 g/mol) = 0.814 grams.

e. If you start with 1.5 g of Na2SiO3 and only 0.814 grams can react with the given 0.800 g of HF, then the amount of Na2SiO3 left after the reaction is (1.5 g - 0.814 g) = 0.686 grams.

Answer:

Half-life of the reaction is 0.072s.

Explanation:

In a first-order reaction, half-life, t1/2, is defined as:

[tex]t_{1/2} = ln2 / k[/tex] (1)

Where k is the rate constant of the reaction.

In the problem, the thermally descomposition of cyclopropane has a rate constant of 9.6s⁻¹. Replacing in (1):

[tex]t_{1/2} = ln2 / 9.6s^{-1}[/tex]

[tex]t_{1/2} = 0.072s[/tex]

I hope it helps!

Answer:

A. Theoretical yield is 36800g

B. 49.73%

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

CO + 2H2 —> CH3OH

From the question given, we obtained the following:

Mass of CO = 37.5 kg = 37.5 x 1000 = 37500g

Mass of H2 = 4.60kg = 4.60 x 1000 = 4600g

Let us convert these Masses to mol

For CO:

Molar Mass of CO = 12 + 16 = 28g/mol

Mass of CO = 37500g

Number of mole of CO = 37500/28 = 1339.3moles

For H2:

Molar Mass of H2 = 2x1 =2g/mol

Mass of H2 = 4600g

Number of mole of H2 = 4600/2 =

2300moles

Now we can see from the equation above that for every 2moles of H2, 1mole of CO is required. Therefore, 2300moles of H2 will require = 2300/2 = 1150moles of CO. This amount(ie 1150moles) is little compared to 1339.3moles of CO calculated from the question. Therefore, H2 is the limiting reactant.

Now we can calculate the theoretical yield as follows:

CO + 2H2 —> CH3OH

Molar Mass of H2 = 2g/mol

Mass of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of CH3OH = 12 + 3 + 16 + 1 = 32g/mol

From the equation,

4g of H2 produced 32g of CH3OH

Therefore, 4600g of H2 will produce = (4600 x 32)/4 = 36800g of CH3OH

Therefore, the theoretical yield is 36800g

B. Actual yield = 1.83 x 10^4g

theoretical yield = 36800g

%yield =?

%yield = Actual yield /Theoretical yield x 100

%yield = 1.83 x 10^4/36800 x 100

%yield = 49.73%

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is [tex]C_5H_7N[/tex]

(b) The empirical formula for the given compound is [tex]C_4H_5N_2O[/tex]

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles[/tex]

Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = [tex]\frac{6.17}{1.23}=5.01\approx 5[/tex]

For Hydrogen  = [tex]\frac{8.70}{1.23}=7.07\approx 7[/tex]

For Nitrogen = [tex]\frac{1.23}{1.23}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is [tex]C_5H_7N_1=C_5H_7N[/tex]

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles[/tex]

Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = [tex]\frac{4.12}{1.03}=4[/tex]

For Hydrogen  = [tex]\frac{5.19}{1.03}=5.03\approx 5[/tex]

For Nitrogen = [tex]\frac{2.06}{1.03}=2[/tex]

For Nitrogen = [tex]\frac{1.03}{1.03}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is [tex]C_4H_5N_2O_1=C_4H_5N_2O[/tex]

Answer:

Shift to the right side

Explanation:

Remember that any stress that we bring to a chemical equilibrium, will be answered by this in such a way as to minimize the disturbance and attain equilibrium again. This is known as LeChatelier´s principle.

If we add N₂(g) to the system at equilibrium

N₂ ( g) + 3 H₂(g)   ⇄ 2 NH₃(g)

the equilibrium will consume part of the added N₂ gas until it reaches equilibrium again producing more ammonia (NH₃).

Mathematically this is equivalent to a  decrease the pressure of N₂ so that the equilibrium partial pressures of the gases obey the equilbrium pressure constant Kp.

Therefore the equilibrium will shift to the product side.

Final answer:

Adding N2(g) to the equilibrium system shifts the equilibrium towards the formation of more NH3(g), as the system seeks to counteract the increase in N2(g) concentration by consuming it, along with H2(g), to produce additional NH3(g).

Explanation:

The question asks about the effect of adding N2(g) to the equilibrium system of the reaction N2(g) + 3 H2(g) ⇆ 2 NH3(g), which is exothermic. According to Le Chatelier's Principle, if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system will adjust itself to counteract that disturbance and a new equilibrium will be established. In this case, adding more N2(g) shifts the equilibrium towards the products to reduce the disturbance, thereby increasing the concentration of NH3(g) and using up some of the added N2(g) along with H2(g) to reestablish equilibrium. This movement towards the products is a direct response to the increased concentration of a reactant, demonstrating the system's attempt to maintain equilibrium.

Using Henry's Law, the solubility of CO2 at 1.8 bar and 298 K is found to be 2970 g/m3. Since there is 1 L of water, this equates to 2.97 g of CO2 dissolved in the 1.00 L bottle of carbonated water.

The question asks how many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at a pressure of 1.8 bar and 298 K using the Henry's Law constant for CO2. To find the concentration of CO2 in the water, we use Henry's Law which states that the solubility of a gas in a liquid is proportional to the pressure of that gas above the liquid. The Henry's Law equation is Sg = kH * Pg, where Sg is the solubility of the gas, kH is the Henry's Law constant and Pg is the partial pressure of the gas.

The density of water is given as 998 kg/m3, which we use to convert the volume of water to mass. Since 1 L of water has a mass of approximately 998 g and the Henry's Law constant for CO2 at 298 K is 1.65×103 bar, we can plug in our values: Sg = (1.65×103 bar)(1.8 bar) = 2970 g/m3 of CO2. Since we have 1 L of water, or 0.998 kg, we can convert to grams: 0.998 kg × 1000 g/kg = 998 g.

To find the grams of CO2 dissolved, we use the solubility in g/m3: 2970 g/m3 × 0.001 m3 = 2.97 g of CO2.

Answer: The net cell equation for the given electrochemical cell is given below.

Explanation:

The given chemical cell follows:

[tex]Co(s)|Co^{2+}(aq.,0.0155M)||Ag^{+}(aq,3.50M)|Ag(s)[/tex]

Oxidation half reaction: [tex]Co(s)\rightarrow Co^{2+}(aq,0.0155M)+2e^-[/tex]

Reduction half reaction: [tex]Ag^{+}(aq,3.50M)+e^-\rightarrow Ag(s)[/tex]       ( × 2)

Net cell reaction: [tex]Co+2Ag^{+}(aq,3.50M)\rightarrow Co^{2+}(aq,0.0155M)+2Ag(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the net cell equation for the given electrochemical cell is given above.

Explanation:

Answer:

There is no short answer.

Explanation:

In the given example Bobby is creating a solution for his bacteria count which consists of 1% bacterial sample.

Considering that the solution was mixed homogeneously, he can apply the procedure to the remaining sample and get the results he wants.

Or if the average number of bacteria in a 1 mL sample is known, he can apply that information proportionally to the 100 mL mixture and find the original cell concentration.

I hope this answer helps.

This is an incomplete question, here is a complete question.

Hydrogen peroxide decomposes to water and oxygen according to the following reaction:

[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]

It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate at certain intervals.

Initial rate determinations at 40°C for the decomposition give the following data:

[H₂O₂] (M)      Initial Rate (mol/L min)

 0.10                  1.93 × 10⁻⁴

 0.20                 3.86 × 10⁻⁴

 0.30                 5.79 × 10⁻⁴

Calculate the half-life for the reaction at 40°C?

Answer : The half life for the reaction is, 3590.7 minutes

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]

Rate law expression for the reaction:

[tex]\text{Rate}=k[H_2O_2]^a[/tex]

where,

a = order with respect to [tex]H_2O_2[/tex]

Expression for rate law for first observation:

[tex]1.93\times 10^{-4}=k(0.10)^a[/tex] ....(1)

Expression for rate law for second observation:

[tex]3.86\times 10^{-4}=k(0.20)^a[/tex] ....(2)

Expression for rate law for third observation:

[tex]5.79\times 10^{-4}=k(0.30)^a[/tex] ....(3)

Dividing 2 by 1, we get:

[tex]\frac{3.86\times 10^{-4}}{1.93\times 10^{-4}}=\frac{k(0.20)^a}{k(0.10)^a}\\\\2=2^a\\a=1[/tex]

Thus, the rate law becomes:

[tex]\text{Rate}=k[H_2O_2]^1[/tex]

[tex]\text{Rate}=k[H_2O_2][/tex]

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

[tex]1.93\times 10^{-4}=k(0.10)^1[/tex]

[tex]k=1.93\times 10^{-3}min^{-1}[/tex]

Now we have to calculate the half-life for the reaction.

The expression used  is:

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]1.93\times 10^{-3}min^{-1}=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=3590.7min[/tex]

Thus, the half life for the reaction is, 3590.7 minutes

Answer:

Volume at half equivalence point, that is 19.19 mL/2 = 9.595 mL.

Explanation:

From the question we are asked to find the volume that would one can use the pH to determine the pKa of the acid and the answer will be the volume at half equivalence point.

At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. And this concept is what is known as the buffer solution.

HA <-----------------------------> H^+ + A^-

(Note: the reaction above is a reversible reaction. Also, the concentration of HA is equal to the concentration of A^-).

Therefore, we can calculate our pka from the equation below(assuming the pH is given).

pH= pka + log ( [A^-] / log [HA].

===> At half equivalence point pH= pKa.

Recall that, pka = - log ka.

Then, ka = 10^-pka.

Where pH= pKa.

Therefore, ka = 10^-pH.

Volume at half equivalence point = 9.595 mL.

At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. It is also known as buffer solution.

Chemical reaction:

[tex]HA--->H^++A^-[/tex]

Calculation of pKa:

[tex]pH= pka + log\frac{[A^-]}{[HA]}[/tex]

At half equivalence point, pH= pKa.

We know, [tex]ka = 10^{-pka}[/tex]

Thus, [tex]ka = 10^{-pH}[/tex] (since, pH= pKa )

Volume is given which is 19.19mL. So, the volume that is used to determine pKa of the acid will be:

19.19/2 = 9.959mL

Find more information about Equivalence point here:

brainly.com/question/24584140

Answer: Part A answer is 180.15588. Part B answer is the moles of HCL required to react with 10 grams of calcium hydroxide.

Explanation:

I just calculated my equation and got my answer.

Answer:

a. 0.076mole

b. 230.88g

Explanation:

a. We'll begin by calculating the molar mass of C6H12O6. This is illustrated below:

Molar Mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol

Mass of C6H12O6 = 13.6g

Number of mole of C6H12O6 =?

Number of mole = Mass /Molar Mass

Number of mole of C6H12O6 = 13.6/180

Number of mole of C6H12O6 = 0.076mole.

Therefore, 0.076mole of C6H12O6 is present in 13.6g of C6H12O6.

b. Let us calculate the molar mass of CaCl2. This is illustrated below:

Molar Mass of CaCl2 = 40 + (35.5x2) = 40 + 71 = 111g/mol

Number of mole of CaCl2 = 2.08 moles.

Mass of CaCl2 =?

Mass = number of mole x molar Mass

Mass of CaCl2 = 2.08 x 111

Mass of CaCl2 = 230.88g

Therefore, 230.88g is required.

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

. A solution that is 0.10 M NH3 and 0.10 M NH4Cl

Explanation:

A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.

A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.

Answer:

Good buffer systems are:  

A) NH3 + NH4Cl

C) HCN + LiCN

D) HF + NaF

Explanation:

Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:

1) Weak acid + salt

or

2) Weak alkaly + salt

It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.

Then, when we evaluate all options in this exercise, answers are the following:

A) 0.10 M NH3 and 0.10 M NH4Cl. It is a buffer because NH3 (ammonia) is a weak alkaly and NH4Cl is a salt coming from NH3.

Buffer component reactions:

Reaction weak alkaly:   NH3 + H2O <-----> NH4+ + OH-

Reaction salt in water:  NH4Cl ---> NH4+ + Cl-

NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Cl is a salt from NH3.

C) 0.10 M HCN and 0.10 M LiCN. It is a buffer because HCN is a weak acid and LiCN is a salt which is coming from HCN.

Buffer component reactions:

Reaction weak acid:     HCN + H2O <-----> H3O+ + CN-  

Reaction salt in water:  LiCN --> Li+ + CN-

CN- is the anion of the acid, so it must be part of the salt in the buffer system. Then LiCN is a salt from HCN.

D) 0.10 M HF and 0.10 M NaF. It is a buffer because HF is a weak acid and NaF is a salt which is coming from HF.

Buffer component reactions:

Reaction weak acid:      HF + H2O <------> H3O+ + F-

Reaction salt in water:   NaF ---> Na+ + F-

F- is the anion of the weak acid (HF), so it must be part of the salt in th buffer systema. Then NaF is a salt coming from HF.

However option B, it is not a buffer, because it is a mixture of 0.10 M HCN and 0.10 M NaF.    Salt is not coming from the weak acid.

Reaction weak acid:    HCN + H2O <-----> H3O+ + CN-  (anion of the acid is CN-)

Rection salt in water:   NaF --> Na+ + F-  (anion in the salt is F-, not CN-)

Anion of the acid is CN- and the anion in the salt is F- so it is not a salt coming from the weak acid. Then option B it is not a buffer system.

Final answer:

ΔH∘rxn is equal to ΔH∘f of the product(s) for the reactions CO(g) + 1/2O2(g) → CO2(g) and 2Li(s) + F2(g) → 2LiF(s).

Explanation:

In chemistry, the enthalpy change of a reaction (ΔH°rxn) is equivalent to the enthalpy of formation (ΔH°f) of the product(s) when the reaction involves the formation of a compound from its elements in their standard states.

The reactions for which ΔH∘rxn is equal to ΔH∘f of the product(s) are:

For these reactions, the enthalpy change of the reaction is equal to the standard enthalpy of formation of the product(s). The enthalpy change of the reaction can be calculated using Hess's law and the enthalpy changes of the individual reactions.

Answer:

Equilibrium Concentration of H₂O(g)  = 0.803

Equilibrium Concentration of Cl₂O(g)  = 0.540

Equilibrium Concentration of HOCl (g) = 0.214

Explanation:

Given;

                H₂O(g)  +   Cl₂O(g) <-------> 2HOCl (g)

I                 0.577        0.314                   0.666                

C               - x              -x                        +2x

E             0.577 - x     0.314 - x               0.666 +2x

[tex]K_c = \frac{[HOCL]^2}{[H_2O][CL_2O]} \\\\0.09 = \frac{[0.666+2x]^2}{[0.577-x][0.314-x]}\\\\0.09(0.1812 -0.891x+x^2) = (0.666+2x)(0.666+2x)\\\\0.0163-0.0802x+0.09x^2 = 0.4436+2.664x+4x^2\\\\3.91x^2+2.7742x+0.4273 =0\\\\x = -0.226, or -0.483[/tex]

Equilibrium Concentration of H₂O(g) = 0.577 - (- 0.226) = 0.803

Equilibrium Concentration of Cl₂O(g) = 0.314 - (- 0.226) = 0.540

Equilibrium Concentration of HOCl (g)  = 0.666 +2(- 0.226) = 0.214

Thus, from the result it can be seen that at equilibrium, the reactants are favored.