Which functional group is used in other functional groups?
A. Ester
B. Carbonyl
c. Hydroxyl
D. Amino

q9​

Answers

Answer 1

Answer:

The answer is B. Carbonyl

Answer 2

Carbonyl is the functional group is used in other functional groups. Therefore, option (B) is correct.

What do you mean by carbonyl functional group?

A functional group with the formula C=O that is composed of a carbon atom double-bonded to an oxygen atom and is divalent at the C atom is known as a carbonyl group in organic chemistry.

A carbonyl gathering is a synthetically natural utilitarian gathering made out of a carbon iota twofold clung to an oxygen molecule - - > [C=O] The most straightforward carbonyl gatherings are aldehydes and ketones typically connected to another carbon compound.

A functional group with a carbon double bonded to an oxygen is called a carbonyl group. They have unsurprising properties, like extremity, instability, and reactivity.

Learn more about carbonyl functional group:

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Related Questions

Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.

Answers

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Write the separation scheme for the isolation of triphenylmethanol from the reaction mixture once the reaction is complete. The separation begins after the addition of HCl and water to the reaction and includes the column chromatography procedure to further purify crude triphenylmethanol isolated in the day 1 procedure.

Answers

````````````````````

What is the concentration (M) of Ch3OH a solution prepared by dissolving of CH3OH sufficient water to give exactly 230 of solution?

Answers

Answer:

1.59 M

Explanation:

What is the concentration (M) of CH₃OH a solution prepared by dissolving 11.7 g of CH₃OH sufficient water to give exactly 230 mL of solution?

Step 1: Given data

Mass of CH₃OH: 11.7 gVolume of solution: 230 mL (0.230 L)

Step 2: Calculate the moles corresponding to 11.7 g of CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

11.7 g × 1 mol/32.04 g = 0.365 mol

Step 3: Calculate the molarity of the solution

M = moles of solute / liters of solution

M = 0.365 mol / 0.230 L = 1.59 M

why beta carbon hydrogen is easily replaced​

Answers

Answer:

γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.

Explanation:

that's all

γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.

A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours

Answers

Answer:

The correct solution is "3.32 gm/L".

Explanation:

Given:

Rate constant,

[tex]K = 0.05 \ hr^{-1}[/tex]

Time,

[tex]t = 24 \ hours[/tex]

Concentration of ethanol,

[tex]C_o= 1 \ mg/L[/tex]

Now,

The concentration of ethanol after 24 hours will be:

⇒ [tex]C_o=C\times e^{-K\times t}[/tex]

By putting the values, we get

    [tex]1=C\times e^{-0.05\times 24}[/tex]

    [tex]1=C\times 0.30119[/tex]

    [tex]C= 3.32 \ gm/L[/tex]

5. How many grams of tin metal can be produced from smelting (heating) of a 4.5 kilograms of tin (IV) oxide? (Note: Elemental tin and oxygen gas are the only products of this reaction).

Answers

Answer:

About 3500 grams of tin.

Explanation:

We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.

First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.

By smelting it, we acquire elemental tin and oxygen gas. Hence:

[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]

(Note: oxygen is a diatomic element.)

The equation is balanced as well.

To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.

Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:

[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]

Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:

[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]

We can convert this into moles.

Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:

[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]

With SnO₂ in the denominator to simplify units.

Finally, we can convert from moles Sn to grams Sn using its molar mass:

[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

With the initial value and above ratios, we acquire:

[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]

Cancel like units:

[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]

Multiply. Hence:

[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]

Since we should have two significant figures:

[tex]=3500 \text{ g Sn}[/tex]

So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.

Answer:

3546g

Explanation:

start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction

SnO2 -> Sn + O2

Sn molecular wt: 119

O2 molecular wt: 32

SnO2 molecular wt:  119+32 = 151

so Sn / SnO2 wt ratio = 119 / 151

4.5 kilograms of tin (IV) oxide will produce:

= 4.5 * 119 / 151

= 3.546 kg

or 3546 grams of tin metal

no need to involve moles ;)

Name the following molecule

Answers

Answer:

It is a Biological Molecule

Based on the "Reactivity in Substitution Reactions" experiment, which molecule would be expected to react the fastest using AgNO3 in water-ethanol ?

Answers

Answer:

C) EtOH 1% AgNO3


The diagram above shows the hydrides of groups 14, 15, 16, and 17 elements. Why does H20, HF, and NH3 have much higher boiling points than the rest of the molecules in their groups?

Answers

Answer:

Hydrogen Bonding

Explanation:

Hydrogen Bonding occurs when a hydrogen atom is bonded to N, O, and F atoms.

The molecules H₂O, HF, and NH₃ all experience hydrogen bonding, which is a relatively strong IMF, causing the molecules to have stronger attraction to each other. Having a stronger attraction between molecules results in more energy required to separate them, thus these molecules will have a higher boiling point than the rest of the molecules in their group.

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data :

ΔH∘rxn 180.5kJ/mol
ΔS∘rxn 24.80J/(mol⋅K)

Required:
a. Calculate the temperature in kelvins above which this reaction is spontaneous
b. Calculate the equilibrium constant for the following reaction at room temperature, 25°C

Answers

Answer:

a. 7278 K

b. Kc = 4.542 × 10⁻³¹

Explanation:

a.

The reaction is spontaneous when ΔG° < 0. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

Then, the reaction will be spontaneous when,

ΔH° - T × ΔS° < 0

T > ΔH°/ΔS

T > (180.5 × 10³ J/mol)/(24.80J/mol⋅K)

T > 7278 K

b.

First, we will calculate ΔG° at 25 °C (298 K)

ΔG° = ΔH° - T × ΔS°

ΔG° = (180.5 × 10³ J/mol) - 298 K × (24.80J/mol⋅K) = 1.731 × 10⁵ J/mol

Then, we will calculate the equilibrium constant (Kc) using the following expression.

ΔG° = - R × T × ln Kc

-ΔG°/R × T =  ln Kc

-(1.731 × 10⁵ J/mol)/(8.314 J/mol.K) × 298 K =  ln Kc

Kc = 4.542 × 10⁻³¹

how does lead resemble chromium?​

Answers

Lead resembles chromium as they both are what you call heavy metals this refers to any metallic chemical element that has a relatively high density examples of heavy metals will include lead,chromium

Classify each of the reactions listed below as a single-displacement, double-displacement, synthesis,
decomposition, oxidation reduction or combustion reaction.
Reaction Type
: 2Na + Cl2 → 2NaCl
: C2H4 + 3O2 → 2CO2 + 2H2O
: 2Ag2O-> 4Ag + O2
: BaCl2 + Na2SO4->BaSO4 +2NaCl
: 2AI + Fe2O3-> 2Fe + Al2O3

Answers

1. Synthesis
2. Combustion
3. Decomposition
4. Double Replacement
5. Single replacement

Consider the reaction: CaCO3(s)CaO(s) CO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions.

Answers

Answer:

the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K

Explanation:

Given the data in the question;

CaCO₃(s)       →     CaO(s)     +     CO₂(g)  

1.58 moles        1.58 moles     1.58 moles

Since 1 mole of CaCO₃ gives 1 mole of CaO and 1 mole of CO₂

Thus, 1.58 mole of CaCO₃ gives 1.58 moles of CaO and 1.58 moles of CO₂.

Now,

At 298 K, standard entropy values are;

ΔS° ( CaCO₃ ) = 92.9 J/mol.K

ΔS° ( CaO )     = 39.8 J/mol.K

ΔS° ( CO₂ )      = 213.7 J/mol.K

So,

ΔS°[tex]_{system[/tex] = ∑ΔS°( product ) - ∑ΔS°( reactant )

ΔS°[tex]_{system[/tex] = [ ΔS°(CaO) + ΔS°( CO₂ ) ] - ΔS°( CaCO₃ )

we substitute

ΔS°[tex]_{system[/tex] = [ 39.8 J/mol.K + 213.7 J/mol.K ] - 92.9 J/mol.K

ΔS°[tex]_{system[/tex] = 160.6 J/mol.K

i.e, for 1 mol CaCO₃, ΔS°[tex]_{system[/tex] = 160.6 J/mol.K

Now, for 1.58 mol CaCO₃,

ΔS°[tex]_{system[/tex] = 1.58 mol × 160.6 J/mol.K

ΔS°[tex]_{system[/tex] = 253.748 J/K

Therefore, the entropy change for the system when 1.58 moles of CaCO3(s) react at standard conditions is 253.748 J/K

What is the oxidation state of nitrogen in N ?

Answers

Answer:

+5

Explanation:

10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.
The resulting solution boils At temperature of 81.20oC. Find the molar mass of solute.
Given that the BP of pure benzene is 80.10oC and Its elevation boiling point constant = 2.53 oC/m.

Answers

Answer: The molar mass of solute is 115 g/mol.

Explanation:

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]

OR

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]

Boiling point of solution = [tex]81.20^oC[/tex]

i = Vant Hoff factor = 1 (for non-electrolytes)

[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute = 10 g

[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 200 g

Putting values in equation 1, we get:

[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]

Hence, the molar mass of solute is 115 g/mol.

  A solution is made by dissolving 20 ml of acetic acid in 180ml of water. Calculate its volume concentration​

Answers

Answer:

water is 9/10

chemical would be 1/10

Explanation:

180/200 would be water concentration in solution

and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)

Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)

Answers

Answer:

Rock A will have far more chemical weathering than Rock B due to the rise in area effect

Explanation:

Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.

1. Why is it necessary to equalize the pressure(i.e, have the water level the same in each tube) before taking a volume reading?

2. Why is it important to use water that has been pre-saturated with CO2 in the gas burettes?

3.If your antacid sample had been contaminated by moisture, what effect(if any )would you expect this to have on your result

4.Explain why an'antacid is called as such,what is the role of the NAHCO3 or CACO3 in reactions?

Answers

Answer:

If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.

Explanation:

This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.

A rigid, sealed container that can hold 26 L of gas is filled to a pressure of
5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?

Answers

Answer:

121 °C

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 374 °C

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

NOTE: Volume = constant

Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 374 °C

Initial temperature (T₁) = 374 °C + 273

Initial temperature (T₁) = 647 K

Next, we shall determine the final temperature. This can be obtained as follow:

Initial pressure (P₁) = 5.97 atm

Initial temperature (T₁) = 647 K

Final pressure (P₂) = 3.64 atm

Final temperature (T₂) =?

P₁ / T₁ = P₂ / T₂

5.97 / 647 = 3.64 / T₂

Cross multiply

5.97 × T₂ = 647 × 3.64

5.97 × T₂ = 2355.08

Divide both side by 5.97

T₂ = 2355.08 / 5.97

T₂ = 394 K

Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 394 K

Final temperature (T₂) = 394 – 273

Final temperature (T₂) = 121 °C

Thus, the new temperature is 121 °C

Predict the Normality of H2SO4 if 75 ml of 96.6 % pure H2SO4 added to 425 ml water. The density of H2SO4 is 1.83 g/cm3?

Answers

Explanation:

Normality is one of the concentration terms.

It is expressed as:

[tex]N=\frac{mass of the substance}{equivalent mass}* \frac{1}{volume of solution in L.}[/tex]

The volume of the solution is 425 mL.

Mass of sulfuric acid given is:

[tex]mass=volume * purity* density\\ = 75 mL * 0.966 * 1.83 g/mL\\\\=132.5 grams\\[/tex]

The equivalent mass of sulfuric acid is 49.0g/equivalents

Hence, the normality of the given solution is:

[tex]N=\frac{132.5g}{49.0g/equi.} *\frac{1000}{425mL} \\Normality=6.36N[/tex]

Answer is: 6.36N.

Categorize the following reaction as an acid-base neutralization, precipitation, combination, decomposition, combustion, displacement, or disproportionation reaction.

Ba(C2H3O2)2(aq) + Na2CO3(aq) → BaCO3(s) + 2 NaC2H3O2(aq)

Answers

Answer:

Precipitation

Explanation:

Let's consider the balanced chemical equation between barium acetate and sodium carbonate to form barium carbonate and sodium acetate.

Ba(C₂H₃O₂)₂(aq) + Na₂CO₃(aq) → BaCO₃(s) + 2 NaC₂H₃O₂(aq)

Both products and reactants are salts. But, among the products, barium carbonate is solid. This allows us to classify it as a precipitation reaction.

If 225 cal of heat is added to 15.0 g of 2-propanol at room temperature, 25ºC, the alcohol will reach a final temperature of ________ ºC. The specific heat of 2-propanol is 0.612 cal/(g•ºC).

Answers

Explanation:

I have no idea about this

Flag A solution of the weak acid, HF, and a solution of the strong acid, HCl, have the same pH. Which solution will require the most sodium hydroxide, NaOH, to neutralize

Answers

Answer:

C) Both will require the same amount because the concentrations are equal.

Explanation:

The pH of a solution is defined as:

pH = -log [H+]

This H+ is the ion that reacts with OH- (From NaOH) as follows:

H+ + OH- → H2O

When all H+ reacts, we can say the solution was neutralized.

Now, as both, the solution with the weak acid and the solution with strong acid have the same pH, we can say that their [H+] is the same. Assuming the volume of both solutions is the same:

Both will require the same amount because the concentrations are equal.

The absorption spectrum of neon has a line at 633 nm. What is the energy of this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's constant is 6.626 x 10-34 J·s.)

Answers

Answer:

B

Explanation:

E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.

Choose all the answers that apply. Silicon (Si) has 14 protons and an atomic mass of 28. Silicon has _____. three electron shells 14 electrons 14 neutrons two electron shells 28 electrons

Answers

Answer:

three electron shells

14 electrons

14 neutrons

Explanation:

Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.

Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.

The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.

how does iron I differ from iron II

Answers

Answer:

Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.

If the temperature of a volume of dieal gas ncreases for 100 to 200, what happens to the average kinetic energy of the molecules?

Answers

Answer:

It increases but less than double

Explanation:

As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.

We know, the kinetic energy of an ideal gas is given by :

[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]

where, R = gas constant

            T = absolute temperature

            M = molecular mass of the gas

From the above law, we get

[tex]$V_{avg} \propto \sqrt{T}$[/tex]

Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.

In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]

[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]

[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]

Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]

Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.

Thus, the answer is " It increases but less that double".


please help!

What is the definition of thermal chemistry?

a.The study of change that involves warm objects

b.The study of change that involves heat

c.The study of change that involves cool objects

d.The study of change that involves temperature

Answers

D. That is the correct answer

Enzyme catalyzing breakdown of atp to adp

Answers

Answer:

ATP hydrolase

Explanation:

Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.

The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.

7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
8. Using the data from question 7 what is the molar concentration of KMnO4 ?
10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Answers

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

Other Questions
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