Which species is the conjugate base of H2SO3

Answer:

-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly

hope it will help you

Explanation:

here is the answer to your question

Answer:

[tex]V_2=482.5mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law due to the fact that we are dealing with variable volume, temperature and pressure:

[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]

In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]

Regards!

Answer:

open system

Explanation:

Answer:

Isolated system

Explanation:

An isolated system is one that cannot exchange either matter or energy with its surroundings.

Answer:

Octadecanol: C18H380

Dimethyl sulfoxide: CHOS

Explanation:

Covalent compounds are formed between non-metallic elements

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.

Answer:

Explanation:

ethyl 3-methylbenzoate

Answer:

the entropy of the universe increases

Explanation:

The second law of thermodynamics can be stated in terms of entropy, in this statement of the law; ''a spontaneous process increases the entropy of the universe''.

Given that; ∆Suniverse = ∆Ssystem + ∆Ssurroundings, the definition of a spontaneous process is one in which ∆Suniverse >0.

Hence, a process is spontaneous when the entropy of the universe increases.

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.

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.

KAnswer:

See explanation

Explanation:

It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.

Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.

Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.

Note that HCl is not a dehydrating agent.

Answer:

negative

the chage on hydroxide ions os negative

Answer:

Anhydrous iron(III) chloride may be prepared by treating iron with chlorine:[11]

{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}

Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.

Dissolving iron ore in hydrochloric acid

{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}

Oxidation of iron(II) chloride with chlorine

{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}

Oxidation of iron(II) chloride with oxygen

{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}

Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride. Hydrated iron(III) chloride can be converted to the anhydrous form by treatment with thionyl chloride.[12] Similarly, dehydration can be effected with trimethylsilyl chloride:[13]

{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}

Anhydrous iron(III) chloride may be prepared by treating iron with chlorine.

Iron filings are small shavings of ferromagnetic material.

[tex]{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}[/tex]

Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.

Dissolving iron ore in hydrochloric acid.

Oxidation of iron(II) chloride with chlorine.

[tex]{\displaystyle {\ce {2FeCl_2_{(}aq){+~}Cl_2_{(}g)- > 2FeCl_3_{(}aq)}}}\\[/tex]

Oxidation of iron(II) chloride with oxygen.

Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride.

Hydrated iron(III) chloride can be converted to an anhydrous form by treatment with thionyl chloride. Similarly, dehydration can be affected by trimethylsilyl chloride.

[tex]{\displaystyle {\ce {FeCl_3.6H2O + 12 Me_3SiCl - > FeCl3 + 6 (Me_3Si)2O + 12 HCl}}}[/tex]

Learn more about the iron filings here:

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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 ;)

Answer:

[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to set up the required chemical reaction as follows:

[tex]C_{14}H_8O_2+O_2\rightarrow CO_2+H_2O[/tex]

However, it turns out mandatory to balance it according to the law of conservation of mass, which states that the atoms must the same on the reactants side as on the products side; thus, we proceed as follows:

[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]

In order to have 14 C atoms, 8 H atoms and 32 O atoms on both sides of the chemical equation.

Regards!

Answer:

I believe it might be blood exiting the lungs, not positive though.

Explanation:

i think it is blood exiting the lungs

Answer:

C₂H₆(g) + O₂(g) ⇒ CO₂(g) + H₂O(l)

Explanation:

Which of the following are combustion reactions?

NaNO₃(aq) ⇒ Na⁺(aq) + NO₃⁻(aq)

NO. This is a dissociation reaction.

C₂H₆(g) + O₂(g) ⇒ CO₂(g) + H₂O(l)

YES. This is a combustion reaction because a compound reacts with oxygen to form carbon dioxide and water.

Mg(s) + HCl (aq) ⇒ H₂(g) + MgCl₂(aq)

NO. This is a single displacement reaction.

HCl(aq) + NaOH (aq) ⇒ HOH(l) + NaCl(s)

NO. This is a neutralization reaction.

Answer:

8.88 grams of aluminum nitrate should be weighted.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to use the definition of molarity to calculate the moles of aluminum nitrate as follows:

[tex]M=\frac{n}{V}\\\\n=M*V\\\\n=0.139mol/L*0.300L=0.0417mol[/tex]

Now, since the molar mass of aluminum nitrate is 212.996 g/mol, we obtain the following mass:

[tex]m=0.0417mol*\frac{212.996 g}{1mol} \\\\m=8.88g[/tex]

Therefore, 8.88 grams of aluminum nitrate should be weighted.

Regards!

here's the answer to your question

The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.

Answer:

The correct option is C ( the activation energy is lowered, the reactant molecules collide more frequently and with greater energy per collision).

Explanation:

The rate of a chemical reaction is defined as the quantity of products that are formed per unit time. Rates can be computed based on either how the reactants are used up or how products are formed.

There are factors that affects the rate of a chemical reaction and they include TEMPERATURE, catalyst, surface area of reacting substances and many among others.

TEMPERATURE increase in most chemical reactions increases the rate of the reaction. This is because molecules gain more energy at higher temperatures. This increases their kinetic energy resulting in more effective collision of the reactant molecules. The rate of reaction depends on the frequency of this effective collisions between the reacting particles.

Effective collision are those that result in reactions, which when they occur, the colliding particles become activated with increased kinetic energy. This kinetic energy must exceed a particular energy barrier for a particular reaction if the reaction must take place. This energy barrier that must be overcome is known as the ACTIVATION ENERGY.

As a result of the time and temperature data collected for the reaction of KMnO4 with either malonic acid or oxalic acid, the reactant molecules collide more frequently and with greater energy per collision.

It is very important to known that the more the reactant molecules is said to collide, the more they will tend to react with one another.

This also leads to a faster reaction rate in any system. So for one to produce a good collision, reactant particles is said  to some minimum amount of energy. This energy, used to start the reaction process, is called the activation energy.

Learn more about activation energy from

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Answer:

trigonal bipyramidal.

Answer:

26kJmole,84kJmole

Explanation:

A catalyst lowers the activation energy between reactants and products. Once this energy barrier is lowered, reactants are converted into products faster.

However, the enthalpy change of a reaction is constant both in the catalysed and uncatalysed reaction.

The activation energy of the uncatalysed reaction must be higher than the activation energy of the catalysed reaction. Hence the answer above.

Answer:

the regulation of Soviet production of missiles that carried nuclear weapons

Explanation:

The aim of the SALT I agreement in 1972 was to stop or greatly reduce the arms race where world powers were stockpiling ballistic missiles and other nuclear arsenal.

Therefore, this agreement brought about the regulation of Soviet production of missiles that carried nuclear weapons

Answer:

D. the regulation of Soviet production of missiles that carried nuclear weapons.

Explanation:

Answer:

molecularity of the rate-determining step

Explanation:

The rate determining step of a sequence of reactions is the slowest step in the sequence of non-elementary reactions.

The molecularity of the slowest step in the reaction mechanism gives us the rate law of reaction.

It is for this cause that the slowest step in the reaction sequence is called ''rate determining step'' since it determines the rate law of reaction.

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".

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)