119. In analytical chemistry, bases used for titrations must often be standardized; that is, their concentration must be precisely determined. Standardization of sodium hydroxide solutions can be accomplished by titrating potassium hydrogen phthalate (KHC8H4O4), also known as KHP, with the NaOH solution to be standardized. a. Write an equation for the reaction between NaOH and KHP. b. The titration of 0.5527 g of KHP required 25.87 mL of an NaOH solution to reach the equivalence point. What is the concentration of the NaOH solution

To calculate the reaction free energy ΔG for this reaction, we need to use the standard free energy of formation values given in a data tab, the stoichiometry of the reaction, and the specific conditions of the reaction, including the concentrations of Pb2+ and Br−. After a series of calculations, we will get the ΔG value in joules, which can be converted to kilojoules.

The task here is to calculate the reaction free energy ΔG for the Pb2+(aq) + 2Br−(aq) = PbBr2(s) reaction at 25.0°C. From the given information, we can start by calculating the number of moles of PbBr2 from its mass. Then, referring to the thermodynamic data tab of the ALEKS, we find the standard free energy of formation (ΔGf°) values for Pb2+(aq), Br−(aq), and PbBr2(s). Now, we can use these values and the definition of ΔG for a reaction in terms of ΔGf° values and stoichiometry.

ΔG = ΣΔGf°(products) - ΣΔGf°(reactants).

Note that the equation must be balanced so each ΔGf° value is multiplied by the stoichiometric coefficient of that substance in the reaction. It is also important to remember to convert the answer to kilojoules if the ΔGf° values are given in joules/mole. Lastly, the concentrations of Pb2+ and Br− are included in the reaction quotient Q to show the reaction's non-standard conditions.

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The reaction free energy under the given conditions[tex]\Delta G \approx -472 \text{ kJ/mol}[/tex].

The reaction free energy for the given chemical reaction can be calculated using the standard free energy of formation for each substance involved in the reaction. The reaction is as follows:

[tex]\[ \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq) \rightarrow \text{PbBr}_2 (s) \][/tex]

The standard free energy change \( \Delta G^\circ \) for the reaction is given by:

[tex]\[ \Delta G^\circ = \Delta G^\circ_f (\text{PbBr}_2) - \left( \Delta G^\circ_f (\text{Pb}^{2+}) + 2 \cdot \Delta G^\circ_f (\text{Br}^-) \right) \][/tex]

Using the thermodynamic data provided in the ALEKS Data tab, we have:

[tex]- \( \Delta G^\circ_f (\text{PbBr}_2) = -260.8 \text{ kJ/mol} \)\\ - \( \Delta G^\circ_f (\text{Pb}^{2+}) = -24.4 \text{ kJ/mol} \)\\ - \( \Delta G^\circ_f (\text{Br}^-) = -121.5 \text{ kJ/mol} \)[/tex]

Plugging in these values, we get:

[tex]\[ \Delta G^\circ = -260.8 \text{ kJ/mol} - \left( -24.4 \text{ kJ/mol} + 2 \cdot (-121.5 \text{ kJ/mol}) \right) \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} + 24.4 \text{ kJ/mol} - 2 \cdot 121.5 \text{ kJ/mol} \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} + 24.4 \text{ kJ/mol} - 243.0 \text{ kJ/mol} \] \[ \Delta G^\circ = -260.8 \text{ kJ/mol} - 218.6 \text{ kJ/mol} \] \[ \Delta G^\circ = -479.4 \text{ kJ/mol} \][/tex]

Relationship:

[tex]\[ \Delta G = \Delta G^\circ + RT \ln(Q) \][/tex]

This reaction is:

[tex]\[ Q = \frac{[\text{PbBr}_2]}{[\text{Pb}^{2+}][\text{Br}^-]^2} \][/tex]

Simplifies to:

[tex]\[ Q = \frac{1}{[\text{Pb}^{2+}][\text{Br}^-]^2} \][/tex]

The initial concentrations are:

[tex]- \( [\text{Pb}^{2+}] = 0.750 \text{ M} \) - \( [\text{Br}^-] = 0.232 \text{ M} \)[/tex]

So:

[tex]\[ Q = \frac{1}{(0.750 \text{ M})(0.232 \text{ M})^2} \] \[ Q = \frac{1}{0.750 \cdot 0.053824 \text{ M}^3} \] \[ Q = \frac{1}{0.040368 \text{ M}^3} \] \[ Q \approx 24.77 \][/tex]

Calculation:

[tex]\[ \Delta G = -479.4 \text{ kJ/mol} + (8.314 \text{ J/(mol·K)} \cdot 298.15 \text{ K}) \cdot \ln(24.77) \] \[ \Delta G = -479.4 \text{ kJ/mol} + (2477.57 \text{ J/mol}) \cdot \ln(24.77) \] \[ \Delta G = -479.4 \text{ kJ/mol} + (2477.57 \text{ J/mol}) \cdot 3.183 \] \[ \Delta G = -479.4 \text{ kJ/mol} + 7896.5 \text{ J/mol} \] \[ \Delta G = -479.4 \text{ kJ/mol} + 7.8965 \text{ kJ/mol} \] \[ \Delta G \approx -471.5 \text{ kJ/mol} \][/tex]

Rounding to the nearest kilojoule:

[tex]\[ \boxed{\Delta G \approx -472 \text{ kJ/mol}} \][/tex]

This is the reaction free energy under the given conditions.

The answer is: [tex]\Delta G \approx -472 \text{ kJ/mol}[/tex].

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = [tex]a=31 cm^2[/tex]

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

[tex]V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3[/tex]

Density of the gold = [tex]d=19.3 g/cm^3[/tex]

[tex]m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g[/tex]

Moles of gold = [tex]\frac{29.915 g}{197 g/mol}=0.152 mol[/tex]

[tex]Au^{3+}+3e^-\rightarrow Au[/tex]

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

[tex]\frac{3}{1}\times 0.152 mol=0.456 mol[/tex] of electrons

Number of electrons = N =[tex]0.456\times \times 6.022\times 10^{23}[/tex]

Charge on single electron = [tex]q=1.6\times 10^{-19} C[/tex]

Total charge required = Q

[tex]Q=N\times q[/tex]

Amount of current passes = I = 8 Ampere

Duration of time  = T

[tex]I=\frac{Q}{T}[/tex]

[tex]T=\frac{N\times q}{I}[/tex]

[tex]=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s[/tex]

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Answer:

Amino acids join by linking the acid group of one amino acid to the amino group of another.

Explanation:

Amino acids are organic molecules that form the basic molecules for making proteins and there are. An amino acid comprises of an acidic carboxyl (-COOH) functional group and an amino group (-NH2)  as well as a side an organic side chain (R group).

In the formation of proteins, several amino acids join together by the formation of peptide bonds between each amino acids to form a long polypeptide. These peptide bods are formed by the linking of the acidic carboxyl group of one amino acid to the amino group of another amino acid, during this process water is removed.

In the tripeptide Gly-Ala-Ser, the amino acid at the N-terminal end is Ser. More than one polypeptide chain may be present in a conjugated protein but not in a simple protein. In solution at physiological pH, the side chain of a polar basic amino acid does not bear a negative charge.

A tripeptide is a chain consisting of three amino acid units. In the tripeptide Gly-Ala-Ser, the amino acid at the N-terminal end is Ser. This is because the N-terminal end is the end of a peptide or protein whose amino group is free, while the C-terminal end has a free carboxyl group. Therefore, statement (1) is true.

Conjugated proteins can consist of more than one polypeptide chain, while simple proteins consist of only one polypeptide chain. Therefore, statement (2) is true.

In solution at physiological pH, the side chain of a polar basic amino acid does not bear a negative charge. Instead, it is positively charged. Therefore, statement (3) is false.

Answer:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

When an area has a net flow of electric charge, an electric current is considered to be present. Electrons traveling via a wire in electric circuits frequently carry this charge.

One or more of the electrons from each atom are only weakly connected to the atom in metals, allowing them to move around freely inside the metal. Electric current is a term used to describe how much electricity flows across a circuit and how it flows in an electronic circuit. Amperes are used to measure it (A). The more electricity flowing across the circuit, the higher the ampere value.

If you imagine electricity as the flow of water in a river, it will be simple to understand. When the electrons collide, the current is the number of electrons flowing per second.

Like current, voltage is a word that is frequently used in relation to electrical circuits. Volts are used to measure voltage (V). Voltage and the movement of electrons in a circuit are connected, just like current and current are. Voltage is the amount of force driving the flowing electrons, whereas the current is the flow of electrons.

Therefore, when there is a net flow of electric charge through an area, an electric current is said to exist. In electrical circuits, electrons traveling over a wire frequently carry this charge.

Read more about electric current, here

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Answer:

hola como estas

Explanation:

Answer:

The molarity of the Br anion is 0.00136 M = 0.0014 M to 2 s.f

Explanation:

Complete full question

A chemist prepares a solution of sodium bromide (NaBr) by measuring out 14. mg of NaBr into a 100 mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Br anions in the chemist's solution. Be sure your answer is rounded to 2 significant digits.

To do this, we first calculate the molarity of the aqueous solution of NaBr.

Molarity = (Concentration in g/L) ÷ (Molar Mass)

(Concentration in g/L)

= (Mass of solute in g) ÷ (Volume of solution in L)

Mass of solute = 14 mg = 0.014 g

Volume = 100 mL = 0.10 L

(Concentration in g/L)

= (Mass of solute in g) ÷ (Volume of solution in L)

(Concentration in g/L) = (0.014/0.1) = 0.14 g/L

Molarity = (Concentration in g/L) ÷ (Molar Mass)

Molar Mass = 102.894 g/mol

Molarity = (0.14/102.894) = 0.0013606236 M = 0.00136 M

Assuming complete dissociation, NaBr dissociates into

NaBr → Na⁺ + Br⁻

1 mole of NaBr gives 1 mole of Br⁻

0.00136 M of NaBr will give 0.00136 M of Br⁻

So, the molarity of the Br anion is 0.00136 M = 0.0014 M to 2 s.f

Hope this Helps!!!

Answer:

C2= 0.16M

Explanation:

C1= 2M, V1= 20ml, C2= ?, V2= 250ml

Applying dilution formula

C1V1= C2V2

2×20 =C2×250

C2= 0.16M

Answer:

Step 1 H2 + 2 NO → N2O + H2O (slow)

step 2 N2O + H2 → N2 + H2O (fast)

Explanation:

It is known that the slowest step in a reaction is the rate determining step in a sequence of reactions (reaction mechanism).

We have two important pieces of information in the question to guide our decision making process.

The overall reaction equation, and the rate expression. The two;

2 H2 + 2 NO → N2 + 2 H2O and rate = k[H2][NO]2 all support the answer given above.

The best matching mechanism to the given rate law is 'H2 + 2 NO → N2O + H2O (slow)' followed by 'N2O + H2 → N2 + H2O (fast)'. This mechanism results in first-order dependence on H2 and second-order dependence on NO, matching the observed rate law.

To find a mechanism that matches the given rate law (rate = k[H2][NO]²), we need to find one where NO is involved in the slow (rate-determining) step twice (which will make the overall reaction second-order with respect to NO), and H2 is involved once (making it first-order with respect to H2). From the proposed mechanisms, we can agree that the first one:

is most likely because the slow step involves one H2 and two NO molecules.

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Answer:

c. 9.94 g

Explanation:

From the question,

Using

mt = m₀e⁻kt.................... Equation 1

Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.

Given: m₀ = 33 g, k = 0.04, t = 30 days.

Substitute into equation 1

mt = 33(e⁻(0.04ˣ30))

mt = 33e⁻¹²/¹⁰

mt = 33/e¹²/¹⁰

mt = 33/3.320

mt = 9.94 g.

Hence the right answer is c. 9.94 g

An explosion in a pressure cooker can be explained by Boyle's Law, which states that pressure and volume are inversely related at constant temperature.So,option B is correct.

The explosion of a pressure cooker while making rice can be explained by Boyle's Law, which relates the volume and pressure of a gas under conditions of constant temperature. According to Boyle's Law, if a gas is compressed to a smaller volume without changing the temperature, the pressure of the gas increases. In a pressure cooker, when the steam cannot escape, the pressure continues to rise, and if the volume is constricted, the cooker may not be able to withstand the increased pressure, leading to an explosion. Therefore, the correct answer to the question is B) Boyle’s Law (relationship between pressure and volume).

Answer:

The element is magnesium

[Ne]3s2

Explanation:

When an atom is excited, electrons move from a lower to a higher energy level. These higher energy levels are called excited states. The ground state is the lowest energy arrangement of electrons.

Excited states are important in spectroscopy. It gives scientists an idea of the unoccupied orbitals in the ground state. This is easily deduced from the fact that the specie has twelve electrons in all.

Magnesium has ground state configuration as shown in the answer but has an excited state as shown in the question.

Answer:

A catalyst increases the rate of reaction by decreasing activation energy. ... Enzymes are highly substrate specific and catalyze reactions by providing an alternate pathway of lower activation energy.

Explanation:

I hope this helps :)

The question given is incomplete because the Ka of the acids were not provided. I got the complete question from google as below:

Three buffers are prepared using equal concentrations offormic acid (HCOOH) and sodium formate, hydrofluoric acid (HF) andsodium fluoride, and acetic acid (CH3COOH)and sodium acetate. Rank the three buffers from highest to lowest pH.

According to the text, the Ka of the acids are as follows:

HCOOH: 1.77 × 10–4

HF: 6.8 × 10–4

CH3COOH: 1.76 × 10–5

Answer:

Based on the Ka values of the acids given, the arrangement of the acids given from the highest to lowest pH is as below:

HF > HCOOH > CH3COOH

Explanation:

For acids, the higher the pH, the higher the pK , also, the lower the pH, the lower the pK.

then

pKa = -log(Ka)

So,

The acid with the highest pH will have the highest Ka value , while the acid with the lowest pH will have the lowest Ka value.

Thus, based on the Ka values of the acids given, the arrangement of the acids given from the highest to lowest pH is as below:

HF > HCOOH > CH3COOH

The buffers are ranked from highest to lowest pH based on the pKa values of the weak acids: acetic acid (highest pH), formic acid, and hydrofluoric acid (lowest pH).

To rank the buffers from highest to lowest pH, we can refer to the pKa values of the corresponding weak acids, since the buffers are prepared using equal concentrations of the weak acids and their conjugate bases. The higher the pKa, the weaker the acid, and therefore the higher the pH of its buffer when the concentrations of the acid and its conjugate base are equal.

The pKa of acetic acid (CH₃COOH) is approximately 4.76.

The pKa of formic acid (HCOOH) is approximately 3.75.

The pKa of hydrofluoric acid (HF) is approximately 3.17.

Therefore, the acetic acid and sodium acetate buffer will have the highest pH, followed by the formic acid and sodium formate buffer, with the hydrofluoric acid and sodium fluoride buffer having the lowest pH.

Answer:

Option A and D

Explanation:

The element with electronic configuration 1s^22s^22p^63s^23p^3 and the element with electronic configuration 1s^22s^22p^3 will show similar chemical properties as they both have the same valence electrons of 5 each. The valence electron of the two elements shows that they both belong to the same group. Elements in the same group naturally have the same chemical properties because they have the same combining power i.e valence electron.

The pair of elements that tend to show the same chemical properties are a and d.

The elements belonging to the same group tend to show the same chemical properties. Based on the electronic configuration, the element having the same number of valence electrons belongs to the same group.

The valence electrons in the given configurations are:

a. [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2\;3p^3[/tex] = 5

b. [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2\;3p^6\;3d^1^0\;4s^2\;4p^5[/tex] = 7

c. [tex]\rm 1s^2\;2s^2\;2p^6\;3s^2\;3p^6[/tex] = 8

d. [tex]\rm 1s^2\;2s^2\;2p^3[/tex] = 5

The element a and d tend to show the same number of valence electrons. Thus both the elements will show the same chemical properties.

The pair of elements that tend to show the same chemical properties are a and d.

For more information about electronic configuration, refer to the link:

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Answer:

[tex]p_2=347.9kPa[/tex]

Explanation:

Hello,

In this case, we use the Gay-Lussac's law which allows us to understand a gas' pressure-temperature behavior as a directly proportional relationship:

[tex]\frac{p_1}{T_1}= \frac{p_2}{T_2}[/tex]

Whereas it is convenient to use the pressure in kPa and the temperature in kelvins in order to compute the required resulting pressure, therefore:

[tex]p_1=55.0psi*\frac{6.89476kPa}{1psi} =379.2kPa\\T_1=30.0+273.15=303.15K\\T_2=5.0+273.15=278.15K[/tex]

Thus, we obtain:

[tex]p_2= \frac{p_1T_2}{T_1}=\frac{379.2kPa*278.15K}{303.15K}\\ \\p_2=347.9kPa[/tex]

Best regards.

Answer:

The pressure in the tire at 5.0 °C is 347.91 kPa

Explanation:

Step 1: Data given

The pressure in a bicycle tire is 55.0 psi

Temperature = 30.0 °C = 303 K

Temperature decreases to 5.0 °C = 278 K

Volume and Amount of moles are held constant

Step 2: Calculate the pressure at the new temperature

P1/T1 = P2 / T2

⇒with P1 = the initial pressure of the bicycle tire is 55.0 psi

⇒with T1 = the initial temeprature = 303 K

⇒with P2 = the pressure at the new temperature

⇒with T2 = the decreased temperature = 278 K

55.0 psi / 303 K = P2 / 278 K

P2 = (55.0 psi / 303 K) * 278 K

P2 = 50.46 psi

Step 3: Convert pressure from psi to kPa

50.46 psi = 50.46 * 6.895 = 347.91 kPa

The pressure in the tire at 5.0 °C is 347.91 kPa

Answer:

0.55mL of carbon tetrachloride

Explanation:

CH4(g) + 2Cl2(g) -------> CCl4(g) + 2H2(g)

From the balanced reaction equation

44800mL of chlorine produces 22400 ml of carbon tetrachloride

If 1.1mL of chlorine were consumed, volume of carbon tetrachloride= 1.1×22400/44800

=0.55mL of carbon tetrachloride

Note: 1 mole of a gas occupies 22.4L volume or 22400mL

Answer: seen below

Explanation:

HCH3CO2 + NaOH --------------> CH3CO2- + H2O

Acid specie- HCH3CO2

base- NaOH

Neutral- Na+

CH3COOH + KOH ----------> CH3COOK + H20

Acid- CH3COOH

Base- KOH

Neutral- K+

Answer:

The question is incomplete. The structures were not added to the question. Find attached of the structure and the given answer.

Explanation:

See the attached file for explanation

Answer:

1.20 V

Explanation:

The standard cell potential is calculated from the expression

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction

The species that will be reduced is the one with the higher standard reduction potential and the species that will be oxidized will be the one with the more negative reduction potential.

Thus for our question we will have

oxidation:

Pb(s)  →   Pb2+(aq) + 2 e-       ε⁰ oxidation       =  -   ε⁰  reduction

                                                                          =   - ( - 0.13 V ) = + 0.13 V

reduction    

Br2(l) + 2 e- → 2 Br-(aq)           ε⁰  reduction     = +1.07 V

ε⁰ cell = ε⁰ oxidation + ε⁰  reduction = + 0.13 V + 1.07 V  = 1.20 V

Answer:

7. Phosphoric acid H₃PO₄ 7.5×10⁻³ 6.2×10⁻⁸ 4.8×10⁻¹³

3. Carbonic acid H₂CO₃ 4.2×10⁻⁷ 4.8×10⁻¹¹

Explanation:

Their blend will result to the closest pH of 8.00

Answer:

b

Explanation:

Answer: The enthalpy of reaction is, -173.2 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.

[tex]5C(s)+6H_2(g)\rightarrow C_5H_{12}(l)[/tex]    [tex]\Delta H=?[/tex]

The intermediate balanced chemical reaction will be,

(1) [tex]C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)[/tex]     [tex]\Delta H_1=-3244.8kJ[/tex]

(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]    [tex]\Delta H_2=-393.5[/tex]

(3) [tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]    [tex]\Delta H_3=-483.5kJ[/tex]

Reversing (1) , Multiply (2) by 5 , (3) by 3 and add

[tex]\Delta H=\Delta H_1+5\times \Delta H_2+3\times \Delta H_3[/tex]

[tex]\Delta H=(+3244.8)+(5\times -393.5)+(3\times -483.5)[/tex]

[tex]\Delta H=-173.2kJ[/tex]

Therefore, the enthalpy of reaction is, -173.2 kJ

Answer:

1. F

2. F

3. F

Explanation:

Determine if the following statements are true or false.

Final answer:

The statements regarding the rate law being written using coefficients from the overall equation and the rate-determining step always being the first step of the reaction are both false. A catalyst does lower activation energy but is not included in the rate law nor in the overall balanced equation.

Explanation:

Determining the rate law of a chemical reaction and the rate-determining step is a critical part of understanding reaction kinetics in chemistry. The first statement, 'The rate law for an overall reaction can be written using the coefficients from the overall reaction,' is false. The rate law cannot be directly inferred from the stoichiometric coefficients in the balanced chemical equation; it is determined empirically and often depends on the mechanism and the slowest, rate-determining step of the reaction.

The second statement, 'The rate-determining step of the reaction is always the first step of the reaction,' is also false. While the rate-determining step can be the first step, this is not always the case. It is the slowest step with the highest activation energy, and not necessarily the first step in the reaction mechanism.

The third statement about a catalyst being a species that lowers the activation energy and shows up in the rate law (most of the time) is partly correct. A catalyst does lower the activation energy and speeds up the reaction but does not appear in the rate law and is not present in the overall balanced equation because it is not consumed in the reaction; thus, this statement is false in the context given.

Answer:

We need 420 cal of heat

Explanation:

Step 1: Data given

Mass of the aluminium = 200.0 grams

Temperature rises with 10.0 °C

Specific heat of aluminium = 0.21 cal/g°C

Step 2: Calculate the amount of heat required

Q =m * c* ΔT

⇒with Q =  the amount of heat required= TO BE DETERMINED

⇒with m = the mass of aluminium = 200.0 grams

⇒with c = the specific heat of aluminium = 0.21 cal/g°C

⇒with ΔT = the change of temperature = 10.0°C

Q = 200.0 grams * 0.21 cal/g°C * 10.0 °C

Q = 420 cal

We need 420 cal of heat (option 2 is correct)

Answer:

420 cal of heat

Explanation:

Answer:

Aldol

Explanation:

The reaction will be an aldol.

Answer: Concentration or pressure of a reactant and temperature

Explanation: The two major factors that can affect the rate of a chemical change are concentration or pressure of a reactant  and temperature.