The complete combustion of ethanol, C2H5OH(l), to form H2O(g) and CO2(g) at constant pressure releases 1235 kJ of heat per mole of C2H5OH.
Write a balanced equation for this reaction.
Express your answer as a chemical equation. Identify all of the phases in your answer.

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

Answer:

(2) Mg²⁺ loses 2 electrons.

Explanation:

Let's consider the following half-reaction.

Mg ⇒ Mg²⁺ + 2 e⁻

Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Answer:

the density if vinegar will also be needed

Explanation:

Because this is an experiment of volumetric analysis

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

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Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]

4. Compute the moles of oxygen by using its molar mass:

[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]

6. Search for the closest whole number (in this case multiply by 2):

[tex]C_3H_6O_2[/tex]

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

[tex]M=12*3+1*6+16*2=74g/mol[/tex]

Which is about three times in the molecular formula, for that reason, the actual formula is:

[tex]C_9H_{18}O_6[/tex]

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

Answer:

2ErF3 + 3Mg → 2Er + 3MgF2

Explanation:

Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.

Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2

D. The name for the set of independent and dependent variables that will be controlled by the scientist.

The statement, that describes the ‘control’ in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."

A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.

In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.

Hence the correct option is D.

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Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

Answer:

+2

Explanation:

Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.

Answer:

Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7

Explanation:

The effective nuclear charge experienced by a 1s electron in helium is +1.70.

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

Answer:

It's harder to melt sodium chloride because of its density and stability.

Explanation:

The bond between Na+ and Cl- is an ionic bond, meaning its an ionic compound. Ionic compounds have more inter molecular forces(forces involving multiple molecules). Chlorine has much less of these forces than NaCl and isn't as dense.

Answer:

It's harder to melt sodium chloride because of its density

Explanation:

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Answer:

8 mol

Explanation:

Step 1: Given data

Moles of copper (II) sulfate: 2 mol

Chemical formula of copper (II) sulfate: CuSO₄

Step 2: Establish the molar ratio of copper (II) sulfate to oxygen

According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.

Step 3: Calculate the moles of O in 2 mol of CuSO₄

[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]

Answer:

See explanation

Explanation:

In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:

Blue carbon

In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:

6+1= 7

For the blue carbon, we will have a Septet.

Red carbons

In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:

2+1=3

For the red carbon, we will have a triplet.

See figure 2

I hope it helps!

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

Answer:

[tex]49.45~%[/tex]

Explanation:

In this case, we have to start with the chemical reaction:

[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]

So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:

[tex](6*12)+(10*1)=82~g/mol[/tex]

With this value we can calculate the grams:

[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]

Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:

[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]

Finally, with these values we can calculate the yield:

[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

Based on the data given, the percentage yield of the student's work is 49.45 %.

The equation of the clay-catalyzed dehydration of cyclohexanol is given below:

From the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.

Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.

First we determine the mass of cyclohexene obtained.

Mass = density * volume

volume of cyclohexene = 500 mL

density = 0.811 g/mL

mass of cyclohexene = 500 * 0.811

mass of cyclohexene = 405.5 g

Number of moles of cyclohexene = mass/molar mass

molar mass of cyclohexene = 82 g

moles of cyclohexene = 405.5 g/82

moles of cyclohexene = 4.945 moles

Percentage yield = 4.945/10 * 100%

Percentage yield = 49.45%

Therefore, the percentage yield of the student's work is 49.45 %.

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Answer:

2. Due to the possession of resonance

Explanation:

In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.

This feature is as a result of resonance.

The correct option is 2.

Answer:

1.22 × 10³ g

Explanation:

Step 1: Write the balanced equation

P₄O₁₀ + 6 H₂O ⇒ 4 H₃PO₄

Step 2: Calculate the moles of H₃PO₄ produced by 3.10 moles of P₄O₁₀

The molar ratio of P₄O₁₀ to H₃PO₄ is 1:4. The moles of H₃PO₄ produced are 4/1 × 3.10 mol = 12.4 mol

Step 3: Calculate the mass corresponding to 12.4 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

[tex]12.4 mol \times \frac{97.99g}{mol} = 1.22 \times 10^{3} g[/tex]

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.