Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals

Answer:

1) before the addition of barium hydroxide

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2)after the addition of barium hydroxide

pH = -log [H⁺] = -log(0.0857) = 1.067

3)at equivalent point, the solution will be neutral

pH = 7.0

4) after adding 40.7mL barium hydroxide

Explanation:

equation of reaction

2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)

1) Before the addition of barium hydroxide

concentration of HBr = 0.276M

[H⁺] = 0.276M

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2) After adding 16.7mL barium hydroxide

moles of [OH⁻] = 16.7mL × 0.105 × 2

=3.507m mol = 3.507 × 10³mol

moles  of [H⁺] = 25.5mL × 0.276M

=7.038m mol = 7.038 × 10³mol

moles of  [H⁺] remaining = (7.038 - 3.421)m mol

= 3.617m mol = 3.617 × 10³mol

[H⁺]= [tex]\frac{3.617}{25.5 + 16.7}[/tex] = 0.0857

pH = -log [H⁺] = -log(0.0857) = 1.067

3) At equivalent point, the solution will be neutral

pH = 7.0

4) After adding 40.7mL barium hydroxide

moles of [OH⁻] = 40.7mL × 0.105M × 2

=8.547

moles of [OH⁻] remaining = 8.547 - 7.038

= 1.509m mol = 1.509 × 10³mol

pOH= -log[OH⁻]= 2.82

pH = 14 - 2.82 = 11.18

Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

[tex]A+B \rightarrow AB\ \ (slow)[/tex]

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

[tex]rate=k[A][B][/tex]

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

Best regards.

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

Answer:

A) linear / linear

Explanation:

In this case, we have a triple bond beetween the atoms (See figure 1). If we have this triple bond we will have an Sp hybridization (in both carbons). We have to remember the relationship between the geometry and the hybridization:

-) Sp3 = Tetrahedral

-) Sp2 = Trigonal

-) Sp = Linear

Due to the hybridization, we will have a linear structure between the atoms. The angle between the atoms is 180º (See figure 2).

So, if we have a hybridization Sp for both carbons, we will have a linear geometry in each carbon. Therefore, the answer is A.

Answer:

53.1% of hydrogen reacts

Explanation:

The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.

The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:

Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water

1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:

0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H

In mass:

0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water

As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:

(Mass that reacts / Mass added) × 100

(0.260g / 0.485g) × 100 =

Answer:

Benzoic acid: 1.288g

Sodium benzoate: 124.48g

Explanation:

Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:

pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)

Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.

As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:

2.5L ₓ (0.35mol / L) = 0.875moles of buffer.

And you can write:

0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)

Replacing (2) in (1)

pH = pKa + log [C7H5O2⁻] / [HC7H5O2]

6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]

1.913 =  log [C7H5O2⁻] / [HC7H5O2]

81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]

81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]

82.846 [HC7H5O2] = 0.875mol

And moles of the benzoate, [C7H5O2⁻]:

[C7H5O2⁻] = 0.875mol - 0.01056mol =

Using molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:

Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g

Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g

Answer:

Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.

HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)

The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.

Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.

Given,

The initial quantity of substance (A₀) = 100

Remaining quantity (At) = 10 - 38 = 62

Time elapse (t) = 407 hours

The rate constant (k) is calculated as:

ln (At ÷ A₀) = - kt

ln (62 ÷ 100) ÷ 407 hour = - k

-0.47803580094 ÷ 407 = - k

k = 0.00117453513

Now, half-life from rate constant (k) is calculated as:

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513

[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours

Therefore, option C. 590 hours is the half-life of the substance.

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Answer:

Answers are in the explanation.

Explanation:

Given concentrations are:

In the reaction:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

Kc is defined as:

Kc = 15 = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are equilbrium concentrations.

Also, you can define Q (Reaction quotient) as:

Q = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are ACTUAL concentrations.

If Q > Kc, the reaction will shift to the left until Q = Kc;

If Q < Kc, the reaction will shift to the right until Q = Kc

If Q = Kc, there is no net reaction because reaction would be en equilibrium.

Replacing with given concentrations:

Answer:

THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE

Explanation:

Using PV = nRT

n = Mass / molar mass

P = 732.6 mmHg = 1 atm = 760 mmHg

So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm

P = 0.964 atm

V = 275 mL = 275 *10 ^-3 L

R = 0.082 Latm/ mol K

T = -28 C = 273 - 28 K = 245 K

mass =  1.95 g

molar mass = unknown

Having known the other variables in the formula, the molar mass of the gas can be obtained.

PV = m R T/ molar mass

Molar mass = m RT / PV

Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3

Molar mass = 39.1755 / 265.1 *10^-3

Molar mass = 39.1755 / 0.2651

Molar mass = 147.78 g/mol

The molar mass of the gas is 147.78 g/mol

Answer:

∆H=  438 KJ/mol

Explanation:

First, we have to find the energy bond values for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the number of bonds:

[tex]Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)[/tex]

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the total energy for each bond:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reagents:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

I hope it helps!

The approximate enthalpy change is:

∆H=  438 KJ/mol

First, we have to find the energy bond values for each compound:

Cl-Cl = 243 KJ/mol

F-F = 159 KJ/mol

F-Cl = 193 KJ/mol

Balanced chemical reaction:

Cl₂ (g) + 3F₂ (g) ⟶ 2ClF₃ (g)

Total number of bond for each:

Cl-Cl = 1

F-F = 3

F-Cl = 6

Total bond energy will be:

Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

F-F = (3*159 KJ/mol) = 477 KJ/mol

F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reactants:

Reactants = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

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Answer:

r = k [X]²

Explanation:

X + Y → Z

Generally, the rate of reaction depends on the concentration of reactants. However, the question stated that the rate depends only on reactant X.

The plot of 1/X versus time giving a straight line signifies that this is a second order reaction.

For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.

From this, our rate law is r = k [X]²

Answer:

a. increase

Explanation:

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.

By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.

Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

Answer:

D

Explanation:

Answer:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

2NO(g) + O₂(g) → 2 NO₂

Explanation:

First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.

3H₂ (g) +  N₂(g)  →  2NH₃ (g)

Afterwards, ammonia reacts to oxygen, to produce NO and H₂O

The equation for the process will be:

4NH₃(g)  + 5O₂(g)  →  4NO(g)  +  6H₂O

Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:

2NO(g) + O₂(g) → 2 NO₂

3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)

Answer:

35%

Explanation:

Percentage yield = actual yield / theoretical yield × 100.

Given:

Actual yield = 63.07g

Theoretical yield = ?

Mole ratio of C7H6O3 to C4H6O3 = 1 : 1

1 mole of C7H6O3 - 138.1g

Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g

∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]

= 35% yield.

Let me know if you found this easy to understand.

Answer:

2284.2 g.

Explanation:

The following data were obtained from the question:

Percentage yield = 47%

Theoretical yield = 4860 g

Actual yield =?

The percentage yield is simply defined as the ratio of actual yield to the theoretical yield multiplied by 100. Mathematically, it is expressed as:

Percentage yield = Actual yield /Theoretical yield x 100

With the above formula, we can obtain the actual yield as follow:

Percentage yield = Actual yield /Theoretical yield x 100

47% = Actual yield /4860

Cross multiply

Actual yield = 47% × 4860

Actual yield = 47/100 x 4860

Actual yield = 2284.2 g

Therefore, the actual yield is 2284.2 g.

Answer:

Chemical reaction B governs the process

Explanation:

The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.

Mathematically;

Number of moles = mass/molar mass

Molar mass of CaCO3 = 100 g/mol

So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles

The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide

Mathematically;

That is same as ;

Number of moles = mass/molar mass

molar mass of CO2 is 44 g/mol

Number of moles of CO2 = 1.13/44 = 0.0256 moles

Now, if we compare the values of these number of moles, we can see that there are almost equal.

What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.

So what we conclude here is that we have an equal mole ratio between the two compounds.

So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate

Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.

Answer:

The Empirical Formular is given as; Ti₆Al₄V

Explanation:

The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.

Elements                        Titanium            Aluminium        Vanadium

Percentage                    64.39                 24.19                   11.42

Divide all through by their molar mass

                                     64.39 / 47.87      24.19 / 27               11.42 / 50.94

                                       =  1.345                = 0.896                 = 0.224

Divide all though  by the smallest number (0.224)

                                     1.345 / 0.224        0.896 / 0.224       0.224 / 0.224

                                     = 6                         = 4                             = 1

The Empirical Formular is given as; Ti₆Al₄V

Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

Titanium :

Divide by Molar mass : = 64.39/47.87 = 1.345

Aluminum :

Divide by Molar mass : = 24.19/27 = 0.896

Vanadium :

Divide by Molar mass : = 11.42/50.94 = 0.224

Divide by the smallest :

Titanium = 1.345 / 0.224 = 6.00

Aluminum = 0.896 / 0.224 = 4

Vanadium = 0.224 / 0.224 = 1

Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

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Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Answer:

[tex]\boxed{\sf Option \ E}[/tex]

Explanation:

All the gases at the same temperature and mass have the same average kinetic energy.

If the masses were different, then the different gases will have different velocities. If the temperature was higher then there would be a greater motion, if the temperature was lower, then there would be less motion.

Answer:

option E

Explanation:

Answer:

D

Explanation:

Double Displacement reaction

Both sides are balanced with option D

The balanced form of the chemical equation shown below is [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex] The correct option is D.

A balanced equation is where the reactant and the product have the number of moles of elements. According to the law, the reaction, and the product have the same number of moles after the reaction, so balancing an equation is important.

To balance an equation, it is significant to see the number of moles of reactant and the same number of moles is in the product side. Here the moles of sodium has to be balanced.

Thus, the correct option is D, [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex]

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Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

Effusion rate of [tex]H_2[/tex],

Molar mass of hydrogen,

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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Explanation:

Answer:

The correct answers are:

c. The concentrations of the products decrease

d. The equilibrium constant decreases

Explanation:

Changes in temperature shift the equilibrium. In this problem, the reaction is endothermic, so it absorbs heat so heat is considered as a reactant:

Reactants + heat ⇒ Products

If the temperature is decreased, the heat is decreased, so reactants are removed from the reaction at equilibrium. According to Le Chaterlier's principle, the system will try to compensate the produced change. If reactants are removed, the systems will form reactants and the equilibrium will shift toward the left (formation of more reactants). In consequence, the amount of products will be decreased.

Thus, acorrect option is: c. The concentrations of the products decrease.

Since the equilibrium constant is given by the ratio of concentration of products over concentration of reactants, if the concentration of products decrease, the equilibrium constant also decreases. So, another correct option is: d. The equilibrium constant decreases.

Answer:

1. 6.92 g of H2O

2i. - 46 KJ of energy.

ii. Option A. Evolved.

Explanation:

1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.

This can be obtained as follow:

The equation for the reaction is given below

2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ

Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.

This is illustrated below:

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Therefore, Xg of H2O will react to 110 kJ of energy i.e

Xg of H2O = (36 x 110)/572

Xg of H2O = 6.92 g

Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.

2i. Determination of the energy.

The balanced equation for the reaction is given below:

CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ

Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.

This is illustrated below:

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

Mass of CO from the balanced equation = 1 x 28 = 28 g

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.

Therefore, - 46 KJ of energy were produced from the reaction.

2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.

Therefore, the heat is evolved.

The question is incomplete, the complete question is;

What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate? more than one answer is possible

A) The release of a water molecule that acts as an electrophile in the reaction with methyl benzoate.

B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.

C)The formation of bisulfate (hydrogen sulfate), which acts as an electrophile in the reaction with methylbenzoate.

D)The release of a water molecule that acts as a nucleophile in the reaction with methyl benzoate.

Answer:

B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.

Explanation:

The benzene ring is known to be stable hence it can only undergo a substitution reaction with the aromatic ring still intact. When the substitution reaction involves an electrophile we refer to the process as electrophillic aromatic substitution. Electrophilic aromatic substitution is a useful synthetic route for many organic compounds.

In the electrophilic substitution of methyl benzoate using the 1:1 volume ratio mixture of H2SO4/HNO3, the nitronium ion (NO2+) is the electrophile generated in the test tube. It is this NO2+ that now reacts with the methyl benzoate to yield the reaction product.

Answer:

V2 = 4.48L

Explanation:

using charles law

V1/T1=V2/T2

3.4/283=V2/373

0.012=V2/373

V2= 0.012 x 373

V2 = 4.48L