Does the number of moles of products increase, decrease, or remain the same when each of the following equilibria is subjected to an increase in pressure by decreasing the volume?
1- CO(g)+H2O(g)⇌CO2(g)+H2(g)
2- 2CO(g)⇌C(s)+CO2(g)
3- N2O4(g)⇌2NO2(g)
_______________________________________________
Ethyl acetate, a solvent used in many fingernail-polish removers, is made by the reaction of acetic acid with ethanol:
CH3CO2H(soln)Aceticacid +C2H5OH(soln)Ethanol⇌CH3CO2H(soln)Ethylacetate+H2O(soln)ΔH∘=−2.9kJ
Part A
Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the temperature is increased?
increase
decrease
Part B
How does Kc change when the temperature is decreased? Justify your answers using Le Chatelier's principle.
How does change when the temperature is decreased? Justify your answers using Le Chatelier's principle.
As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in Kc.
As the temperature is decreased, the reaction shifts from right to left. The product concentrations decrease, and the reactant concentrations increase. This corresponds to an decrease in Kc.
As there is neither products no reactants in gas state the temperature does not shift the reaction. So decrease in temperature does not change Kc.
________________________________________________________
A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide:
2CO(g)+O2(g)⇌Pt2CO2(g)ΔH∘=−566kJ
Suppose that you have a reaction vessel containing an equilibrium mixture of CO(g), O2(g), and CO2(g). Under the following conditions, will the amount of CO increase, decrease, or remain the same after each of the following changes?
A platinum catalyst is added.
increase
decrease
remain the same
The temperature is increased.
increase
decrease
remain the same
The pressure is increased by decreasing the volume.
increase
decrease
remain the same
The pressure is increased by adding argon gas.
increase
decrease
remain the same
The pressure is increased by adding gas.
increase
decrease
remain the same

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.

Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.

CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+

Answer:

Option B. S

Explanation:

All of the options except sulphur, S is metal.

Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.

Sulphur, S has atomic number of 16 with electronic configuration as:

S (16) => 1s² 2s²2p⁶ 3s²3p⁴

From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.

Therefore, sulphur, S will gain two electrons to form ion.

As stated earlier, the rest option given are all metals which will form ion by losing electron(s).

Answer

B) Sulphur (S)

Explanation

Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.

Now, let us understand what is a metal and a non-metal.

So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.

Now let us also look at the electronic configuration of Sulphur to get the picture more clearly

        [tex]S\rightarrow 1s^2\; 2s^2\;2p^6\;3s^2\;3p^4[/tex]

so here the p-subshell is incomplete and is in need of 2 electrons.

Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.

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Answer:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

[tex]Entalpy=-2861.9~KJ[/tex]

Explanation:

In this case, we have to start with the reagents:

[tex]Al~+~NH_4NO_3[/tex]

The compounds given by the problem are:

-) Nitrogen gas =  [tex]N_2[/tex]

-) Water vapor  =  [tex]H_2O[/tex]

-) Aluminum oxide =  [tex]Al_2O_3[/tex]

Now, we can put the products in the reaction:

[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]

When we balance the reaction we will obtain:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

Now, for the enthalpy change, we have to find the standard enthalpy values:

[tex]Al_(_S_)=0~KJ/mol[/tex]

[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]

[tex]N_2_(_g_)=0~KJ/mol[/tex]

[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]

[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]

And the products:

[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]

Finally, for the total enthalpy we have to subtract products by reagents :

[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]

I hope it helps!

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂    

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄  4CO₂ + 10H₂O + 2N₂              

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.  

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

[tex]n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles[/tex]      

[tex]n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles[/tex]

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

[tex]n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles[/tex]

[tex]m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g[/tex]

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

[tex] \% = \frac{R_{r}}{R_{T}}*100 [/tex]

Donde:

[tex]R_{r}[/tex]: es el rendimiento real

[tex]R_{T}[/tex]: es el rendimiento teórico

[tex]\% = \frac{3,5}{5,043}*100 = 69,4[/tex]

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!        

Answer:

Osmolarity of solution of KCI = 0.40 osmol

Explanation:

Given:

KCL ⇒ K⁺ + Cl⁻

Find:

Osmolarity of solution of KCI

When M = 0.20 M

Computation:

1 mole of KCL = 2 osmol

1 M of KCl = 2 Osmolarity

So,

Osmolarity of solution of KCI = 2 × 0.20

Osmolarity of solution of KCI = 0.40 osmol

Answer:

a. 1

b. 0.465M NaOH

Explanation:

KHP reacts with NaOH as follows:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:

1/1 = 1

b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:

15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP

In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.

The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:

0.0093 moles NaOH / 0.020L =

a. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.

The balanced equation for the reaction:

NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O

b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point

Molarity of KHP solution = 0.600 M

Volume of KHP solution = 15.5 mL = 0.0155 L

Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L

Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point

Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L

Molarity of NaOH solution ≈ 0.465 M

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Answer:76 grams

Explanation:

Fe+H₂SO₄-->FeSO₄+H₂

For one mole of Fe we get 1 mole of feso4, therefore for 0.5 moles of Fe we get 0.5 moles of feso4.

The molar mass of feso4 is AFe+AS+4AO(A is atomic mass)

56+32+4*16=152grams/mole

Now, we need to multiply the number of moles by the molar mass to get the mass that reacts

152*0.5=76 grams

Answer:

[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

Explanation:

We can start with the reaction of hydrogen and nitrogen to produce ammonia, so:

[tex]N_2~+~H_2~->~NH_3[/tex]

When we balance the reaction we will obtain:

[tex]N_2~+~3H_2~->~2NH_3[/tex]

Now, the production of nitric acid with oxygen would be:

[tex]NH_3~+~O_2~->~HNO_3~+~H_2O[/tex]

If we balance the reaction we will obtain:

[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]

Now, if we put the reactions together we will obtain:

[tex]N_2~+~3H_2~->~2NH_3[/tex]

[tex]NH_3~+~2O_2~->~HNO_3~+~H_2O[/tex]

We can multiply the second reaction by "2":

[tex]N_2~+~3H_2~->~2NH_3[/tex]

[tex]2NH_3~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

We have "[tex]2NH_3[/tex]" on both sides. In the first reaction is in the right in the second reaction is on the left. Therefore we can cancel out this compound and we will obtain:

[tex]N_2~+~3H_2~+~4O_2~->~2HNO_3~+~2H_2O[/tex]

On this reaction, we will have 2 nitrogen atoms on both sides, 6 hydrogen atoms on both sides, and 8 oxygen atoms on both sides. So, this would be the net reaction for the production of nitric acid.

Answer:

It will take 6486.92 minutes  for [H2O2] to fall from 0.95 M to 0.33 M

Explanation:

The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.

In first order reaction;

[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]

where;

a = concentration at time t

[tex]a_o[/tex] = initial concentration

and k = constant.

[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]

[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]

[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]

t = 6486.92 minutes

B. it means the compound is dissolved in water

Answer:

b

Explanation:

a p e x :)

Answer:

The reaction given in the question is:  

2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)

The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:

ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.  

The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.  

Now the values of ΔH° and ΔS° are,  

ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)

ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)

ΔH°rxn = -1277.56 kJ/mole

ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)

ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15

ΔS°rxn = 313.11 J/mole/K

Now the formula for calculating ΔG°rxn is,  

ΔG°rxn = ΔH°rxn - TΔS°rxn

ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole

ΔG°rxn = -1370.86 kJ/mol

Answer:

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

Explanation:

a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

Hope this Helps!!!

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Calculation for Enantiomeric excess:

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) Calculation of percentage:

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

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Answer: Mass of methyl 4-fluoro-3-nitrobenzoate = 199 mg;

Mass of 4-chlorobenzylamine = 282 mg

Explanation: The mass and mol of a molecule is related by its molar mass, which is given in g/mol.

The molar mass of methyl 4-fluoro-3-nitrobenzoate, which has molecular formula: [tex]C_{8}H_{6}FNO_{4}[/tex] is:

[tex]C_{8}H_{6}FNO_{4}[/tex]  = 12.8 + 6.1 + 19 + 14 + 16.4 = 199 g/mol

Since it is asking in mg: MM = 199.10³mg/mol

For 4-chlorobenzylamine, with molecular formula [tex]C_{7}H_{8}ClN[/tex]:

[tex]C_{7}H_{8}ClN[/tex] = 12.7 + 8.1 + 35 + 14 = 141 g/mol

In mg: MM = 141.10³mg/mol

The reaction requires 1 mmol of [tex]C_{8}H_{6}FNO_{4}[/tex] , then its mass is:

m = 1.10⁻³ mol * 199.10³mg/mol = 199 mg

For [tex]C_{7}H_{8}ClN[/tex], it requires 2mmol:

m = 2.10⁻³ mol * 141.10³ mg/mol = 282 mg

For the SNAr reaction, it is necessary 199 mg of methyl 4-fluoro-3-nitrobenzoate and 282 mg of 4-chlorobenzylamine

The mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

To determine the masses of both reagents,

First we will determine their molar masses

Molar mass = 199.14 g/mol

Now, for the mass of 1 mmol of methyl 4-fluoro-3-nitrobenzoate

Using the formula

Mass = Number of moles × Molar mass

Mass = 1 mmol × 199.14 g/mol

Mass = 199.14 mg

Molar mass = 141.6 g/mol

Now, for the mass of 2 mmol of 4-chlorobenzyl amine

Mass = 2 mmol × 141.6 g/mol

Mass = 283.2 mg

Hence, the mass of 1 mmol methyl 4-fluoro-3-nitrobenzoate is 199.14 mg and the mass of 2 mmol of 4-chlorobenzyl amine is 283.2 mg

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Answer:

P SO₂ = 0.06atm

P O₂ = 2.635atm

P SO₃ = 0.53atm

Kp = 29.6

Explanation:

The reaction of Sulfur dioxide and oxygen react to form sulfur trioxide is as follows:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

And Kp is defined as:

[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]

Where P represents the pressure at equilibrium of each reactant.

If you add, in the first, 0.59atm of SO₂ and 2.9atm of O₂, the equilibrium pressures will be:

P SO₂ = 0.59atm - 2X

P O₂ = 2.9atm - X

P SO₃ = 2X

Where X represents the reaction coordiante.

As equilibrium pressure of SO₃ is 0.53atm:

0.53atm = 2X

0.265atm = X

Replacing, equilibrium pressures of each species will be:

P SO₂ = 0.59atm - 2×0.265atm

P O₂ = 2.9atm - 0.265atm

P SO₃ = 2×0.265atm

And Kp will be:

[tex]Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}[/tex]

[tex]Kp = \frac{0.53^2}{{0.06}^2*{2.635}}[/tex]

Answer:

The osmotic pressure of ocean water is about 27 atm.

Explanation:

Pure water is water that contains no impurities. Ocean water is 96.5% pure with only about 3.5% of its content, salt water.

Osmotic pressure occurs when solutions that have different concentrations are isolated by a membrane. This osmotic pressure makes water move towards the solution that has the highest concentration, which means that if the concentration or temperature of the solution is high, the osmotic pressure becomes higher.

The equation for osmotic pressure is pi = iMRT.

Answer:

NH4Br + AgNO3 —> AgBr + NH4NO3

Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

Answer:

[tex]\triangle G = -911.296 \ kJ[/tex]

Explanation:

ΔG = ΔH-TΔS

Where ΔH = -176 kJ = -176000 J , T = 25°C + 273 = 298 K , ΔS = -284.8 JK⁻¹

=> [tex]\triangle G =-176000 - (298)(-284.8)[/tex]

=> [tex]\triangle G = -176000+84870.4[/tex]

=> [tex]\triangle G = -91129.6 \ J[/tex]

=> [tex]\triangle G = -911.296\ kJ[/tex]

Since the value is negative, the reaction is spontaneous under standard conditions at 298 K and the reactants have more free energy than the products.

Answer:

The given statement is false.

Explanation:

So that the given is incorrect.

Answer:

[tex][OH^-]=0.01165M[/tex]

Explanation:

Hello,

In this case, for the dissociation of methylamine:

[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]

We can write the basic dissociation constant as:

[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]

That in terms of the reaction extent [tex]x[/tex], turns out:

[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]

[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]

That has the following solution for [tex]x[/tex]:

[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]

Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:

[tex][OH^-]=0.01165M[/tex]

Best regards.

Answer:

[tex]\boxed{From \ +6 \ to \ 0}[/tex]

Explanation:

2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g)

In the given reaction, Iron in the reactants side have the oxidation number of +6. This is because [tex]O_{3}[/tex] with [tex]Fe_{2}O_{3}[/tex] has oxidation state -6, So any atom with it would have an oxidation state of +6 to give the resultant of zero.

In the products side, Iron acts as a free element reacting with no other atom. So, as per the rule of oxidation states, the oxidation state of Iron in the products side will be zero.

So, the oxidation number changes from +6 to 0 .

Extra Info: Decrease in oxidation state is Reduction , So Iron is being reduced here.

The change in the oxidation number of the iron atom in the reaction is from +3 to 0

Oxidation is simply defined as the the loss of electron. However, Oxidation number simply talks about the number of electrons that is either gained or lossed during bond formation.

The change in the oxidation number of iron in the reaction can be obtained as follow:

2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g)

Oxidation number of Fe₂O₃ = 0 (ground state)

Oxidation number of oxygen = –2

Fe₂O₃ = 0

2Fe + 3O = 0

2Fe + 3(–2) = 0

2Fe – 6 = 0

Collect like term

2Fe = 6

Divide both side by 2

Fe = 6/2

Thus, the oxidation number of Fe in Fe₂O₃ is +3

Therefore, the change in the oxidation number of the iron, Fe, atom in the reaction is from +3 to 0

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Answer:

D.

Explanation:

If a compound is composed of oppositely charged ions, it has to be formed by metal and non-metal.

Li2O

Li - metal

O - non-metal

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

Answer:

Three isomers of hexane are: 2-methylpentane, 3-methylpentane, and 2,3-dimethylbutane.

They are constitutional isomers because they each contain exactly the same number and type of atoms, in this case, six carbons and 14 hydrogens and no other atoms.

Explanation:

Hope it helps.

Answer:

91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Explanation:

The formula for sodium phosphate is Na₃PO₄

Molar mass of sodium phosphate = 164 g/mol

The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;

Na₃PO₄ ------> 3Na⁺ + PO₄³

Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles  of Na₃PO₄

Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g

Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Answer:

760 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

Answer:

Explanation:

Let the monoprotic acid be HX

HX ⇄ H⁺ + X⁻

pH = 2.53

Hydrogen ion concentration

[tex][ H^+]=10^{-2.53}[/tex]

[tex][ X^-]=10^{-2.53}[/tex]

Concentration of undissociated acid will remain almost the same as it is a weak acid

So

Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid

=  [ H⁺] x [Cl⁻ ] / [ HX]

[tex]k_a=\frac{10^{-2.53}\times 10^{-2.53}}{.0166}[/tex]

[tex]k_a=\frac{.00295^2}{.0166}[/tex]

= 5.24 x 10⁻⁴ M .

Answer: C. Metals tend to easily lose their valence electrons.

Explanation:

Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.

Example: Gold, potassium etc

[tex]M\rightarrow M^++e^-[/tex]

Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.

Example: Sulphur, Chlorine

[tex]N+e^-\rightarrow N^-[/tex]

Answer:

filtration, drying, and weighing

Explanation:

The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.

The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.

The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.

Answer:

0.3 mol

Explanation:

Assuming HCl is in excess and Zn is the limiting reagent,

from the balanced equation, we can see the mole ratio of Zn:H2 = 1:1,

which means, each mole of zinc reacted gives 1 mole of H2.

So, if 0.30 mol Zn is added, the no. of moles of H2 produced will also be 0.3 mol, since the ratio is 1:1.