give an example for photodecomposition reaction​

Answers

Answer 1

A decomposition reaction occurs when one reactant breaks down into two or more products. This can be represented by the general equation: AB → A + B. Examples of decomposition reactions include the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.


Related Questions

Choose the compound that exhibits hydrogen bonding as its strongest intermolecular force.
A. C2H6
B. CH3OH
C. CH2Br2
D. SBr2
E. None of the above compounds exhibit hydrogen bonding.

Answers

Answer:

B

Explanation:

To form hydrogen bondings between the molecules, the compound needs a highly electronegative atom (usually N, O, or F) bonded with a hydrogen atom;

and that the highly electronegative atom has lone pair outermost shell electrons.

In the 5 options, only B (CH3OH) has an N, O, or F atom that has lone pair outermost shell electrons (2 lone pairs on each O atom), so it can form hydrogen bonds within its molecules.

Hydrogen bonds are stronger than the van der Waals' forces between its molecules (that exist regardless of whether there are hydrogen bonds).

The compound that exhibits hydrogen bonding as its strongest intermolecular force is  CH₃OH as electronegative oxygen atom is bonded to hydrogen atom.

What is compound?

Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.

Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:

1)Molecular compounds where in atoms are joined by covalent bonds.

2) ionic compounds where atoms are joined by ionic bond.

3)Inter-metallic compounds where atoms are held by metallic bonds

4) co-ordination complexes where atoms are held by co-ordinate bonds.

They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.

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Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces electrical current spontaneously. c) It consumes electrical current to drive a nonspontaneous chemical reaction.

Answers

Answer:

b) It produces electrical current spontaneously.

Explanation:

Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.

The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.

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

Answers

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

What is it’s molecular formula for C5H4 if it’s molar mass is 128.17g/mol

Answers

Answer :QUESTION ①)

✔ C5H4 has a molecular molar mass of :

M(C5H4) = 5 x M(C) + 4 x M(H)

M(C5H4) = 5 x 12 + 4 x 1 M(C5H4) = 60 + 4 M(C5H4) = 64 g/mol

✔ The molecular mass of C5H4 is therefore 64 g/mol.

But, 128/64 = 2 This is double the molar mass of C5H4, this molecule has the formula 2C5H4.

Answer:

C10H8

Explanation:

I clicked on that answer, and it is correct.

A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution

Answers

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

1L of bleach has a mass of 1,100 grams, 7.25% of the mass of bleach is NaClO, 1 mol of NaClO has a mass of 74.44 grams. What is the molarity (mol/L) of NaClO in the bleach? A.0.097 B.0.93 C. 1.07 D.79.75

Answers

Answer:

C. 1.07 M.

Explanation:

Hello,

In this case, we can define the molarity of the bleach as shown below:

[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]

In such a way, given the mass of bleach in a 1-L solution, we can compute the density:

[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]

In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:

[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]

Afterwards, using the by-mass percent of bleach we compute the mass:

[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]

And the moles:

[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]

Therefore, the molarity turns out:

[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]

Thus, answer is C. 1.07 M.

Regards.

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C upper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.

Answers

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

Answer:

A.

Explanation:

Did it on Edge.

If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?

Answers

The Average atomic mass of phosphorus is 29.9.

What is Average atomic mass ?

The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.

Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.

An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.

The versions of an element with different neutrons have different masses and are called isotopes.

The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,

Average atomic mass of P = ∑(Isotope mass) (its abundance)

∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)

Abundance of isotope = % of the isotope / 100.

∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.

Hence , The Average atomic mass of phosphorus is 29.9.

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Draw the major product(s) obtained when the following compounds are treated with bromine in the presence of iron tribromide.

a. Bromobenzene
b. ortho-Xylene
c. Benzene sulfonic acid
d. Benzaldehyde
e. meta-Nitrotoluene
f, para-Dibromobenzene
g. Nitrobenzene tert-Butylbenzene
h. Benzoic acid
i. Dibromobenzene

Answers

Answer:

The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.

Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.      

HBr will always be the side product of the bromine reactions along with the major compound.                                      

Explanation:

Bayer Villiger Provide a balanced chemical equation of the reaction performed in this experiment. Use structures and compound names to show ALL reactants and products involved. Baeyer-Villiger Reaction of Acetophenone Data Results
• Moles of acetophenone used: (Show calculations) 0.020 moles (2.40g/120.151 g mol-1 =0.0199 moles)
• Moles of mCPBA used: (Show calculations) 0.036 moles_(6.25 grams/ 172.56 g.mol-1)
• Expected mass of the product: (Show calculation. Clearly show the limiting and excess reactants)

Answers

Answer:

See the explanations

Explanation:

In the Baeyer-Villiger reaction, we will produce an ester from a ketone (see the first reaction). In our case, the ketone is Acetophenone therefore phenyl acetate would be produced.

Now, for the mass calculation, we have to keep in mind that we have a reaction with a 1:1 ratio. So, if we have 0.02 moles of acetophenone and 0.036 moles of m-CPBA the limiting reagent would be the smallest value in this case acetophenone.

Additionally, if we have a 1:1 ratio and the limiting reagent is 0.02 moles of acetophenone we will have as product 0.02 of phenyl acetate, if we take into account the molar mass of phenyl acetate (136.05 g/mol), we can do the final calculation:

[tex]0.02~mol~acetophenone\frac{1~mol~phenyl acetate}{1~mol~acetophenone}\frac{136.05~g~phenyl acetate}{1~mol~phenyl acetate}=2.72~g~phenyl acetate[/tex]

I hope it helps!

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 

Answers

Answer:

ΔH = [tex]q_{p}[/tex]

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

[tex]q_{com}[/tex] = [tex]q_{p}[/tex]

where

[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

[tex]q_{com}[/tex]  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = [tex]q_{p}[/tex] + 0

==> ΔH = [tex]q_{p}[/tex]

Amanda is doing a report for her Earth Science class about the four seasons. Which of the following would be an effective scientific model to incorporate in her project? a. A calendar indicating the first days of autumn, winter, spring, and summer b. Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season c. A poster board with pictures of weather characteristic of the four seasons d. A bar graph indicating average high and low temperatures for Amanda’s area in the autumn, winter, spring, and summer Please select the best answer from the choices provided

Answers

Answer:

(B)

Explanation:

edg 2020

The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is  Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.

What are seasons?

Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.

The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.

Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.

Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.

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Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
Al(NO3)3
C2H5NH3NO3
NaClO
RbI
CH3NH3CN

Answers

Answer:

Al(NO₃)₃: Acidic.

C₂H₅NH₃NO₃: Acidic.

NaClO: Basic

RbI: pH-neutral

CH₃NH₃CN: Solution basic

Explanation:

The general rules to determine if a solution is acidic, basic or neutral are:

If it is a salt of a strong acid and base, the solution will be pH-neutral. If it is a salt of a strong acid and a weak base, the solution will be acidic due to the hydrolysis of the weak base component (cation). If it is a salt of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the weak acid component (anion).

For the salts:

Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC

C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.

NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.

RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL

CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and  the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC

Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.

What is pH?

pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.

Al(NO₃)₃ is a salt which is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base Al(OH)₃, so the resultant solution of the salt is acidic in nature.C₂H₅NH₃NO₃ salt is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base C₂H₅NH₃OH, so the resultant solution of the salt is acidic in nature.NaClO is a salt of weak acid is HClO and strong base NaOH, so the resultant solution of the salt is basic in nature.RbI salt is formed by the combination of strong acid HI and strong base RbOH, so the resultant solution of the salt is neutral in nature.CH₃NH₃Cyanide is a salt of weak acid hydrogen cyanide and weak base CH₃NH₃OH,  so the resultant solution of the salt is neutral in nature.

Hence, appropriate differentiation was done above.

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Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium​

Answers

Answer:

Rubidium

Explanation:

A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates (The data from the three experiments is in the table). 1. Write the rate law 2. Solve for k.

Answers

Answer:

1. [tex]Rate =k [NO]^{2}[Cl_{2}][/tex]

2. [tex]k= 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]

Explanation:

[tex]Rate =k [NO]^{m}[Cl_{2}]^{n}[/tex]

[tex]Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2[/tex]

[tex]Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1[/tex]

[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}[/tex]

[tex]Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}[/tex]

The volume of ammonia gas at 1.14 atm of pressure is gradually decreased from 90.3 mL to 43.4 mL. What is the final pressure of ammonia if there is no change in temperature

Answers

Answer:

THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.

Explanation:

At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.

Mathematically written as:

P1V1 = P2V2

Re-arranging the formula by making P2 the subject of the formula;

P2 = P1V1 / T2

P1 = 1.4 atm

V1 = 90.3 mL

V2 = 43.4 mL

P2 = unknown

So therefore, we have:

P2 = 1.4 * 90.3 / 43.4

P2 = 2.91 atm

The final pressure of ammonia is therefore 2.91 atm.

Answer:

2.37 atm

Explanation:

Step 1: Given data

Initial pressure of ammonia (P₁): 1.14 atmInitial volume of ammonia (V₁): 90.3 mLFinal pressure of ammonia (P₂): ?Final volume of ammonia (V₂): 43.4 mLConstant temperature

Step 2: Calculate the final pressure of ammonia

Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.

[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]

A 8.22-g sample of solid calcium reacted in excess fluorine gas to give a 16-g sample of pure solid CaF2. The heat given off in this reaction was 251 kJ at constant pressure. Given this information, what is the enthalpy of formation of CaF2(s)

Answers

Answer:

The enthalpy of formation of CaF₂ is -1224.4 kJ.

Explanation:

The enthalpy of formation of CaF₂ can be calculated as follows:

[tex] \Delta H_{f} = \frac{q}{n_{CaF_{2}}} [/tex]

Where:

q: is the heat liberated in the reaction = -251 kJ

The number of moles of CaF₂ is:

[tex] n_{CaF_{2}} = \frac{m}{M} [/tex]

Where:

m: is the mass of CaF₂ = 16 g

M: is the molar mass of CaF₂ = 78.07 g/mol

[tex] n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles [/tex]

Now, the enthalpy of formation of CaF₂ is:

[tex]\Delta H_{f} = \frac{q}{n_{CaF_{2}}} = \frac{-251 \cdot 10^{3} J}{0.205 moles} = -1224.4 kJ/mol[/tex]

Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.

I hope it helps you!      

if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?

Answers

Answer:

The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of the ions in solution were [Cr3+] = 0.75 M and [Fe2+] = 0.25 M at 298K?

Answers

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

                            E°/V

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

                                               E°/V

2Cr³⁺ + 6e⁻ ⇌ 2Cr;               +0.74

3Fe  ⇌ 3Fe²⁺ + 6e-;              -0.41

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

3Fe  ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

[tex]E = E^{\circ} - \dfrac{RT}{zF}\ln Q[/tex]

(a) Data

  E° = 0.33 V

   R = 8.314 J·K⁻¹mol⁻¹

   T = 298 K

   z = 6

   F = 96 485 C/mol

(b) Calculations:  

[tex]Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}[/tex]

 

what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4

Answers

Answer:

2

Explanation:

First, find the hydronium ion concentration of the solution with a pH of 4.

[H₃O⁺] = 10^-pH

[H₃O⁺] = 10⁻⁴

[H₃O⁺] = 1 × 10⁻⁴

Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.

[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01

Lastly, find the pH.

pH = -log [H₃O⁺]

pH = -log (0.01)

pH = 2

The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.

Hope this helps.

The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. What is the theoretical yield of liquid iron, in grams? Just enter a numerical value. Do not enter units.

Answers

Answer:  313.6

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]

[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]

According to stoichiometry :

1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex]  of [tex]CO[/tex]

Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.

As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]

Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]

Theoretical yield of liquid iron is 313.6 g

Use your periodic table and calculator as needed for the following question.
How much stock solution is needed to make 250 mL of a 6.0M solution. The molarity of the stock solution is 18M.
Selections may be rounded so choose the best answer.
56 mL
83 mL
2.3 mL
4.7 ml

Answers

Hope you find this answer I need points

A chemistry student weighs out of chloroacetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.

Answers

Answer:

11.6mL of the 0.1400M NaOH solution

Explanation:

0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.

The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:

ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.

You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:

0.154g ₓ (1mol / 94.5g) = 1.63x10⁻³ moles of ClCH₂COOH

To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:

1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =

11.6mL of the 0.1400M NaOH solution

A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.

Answers

Answer:

Cu₂S

Explanation:

From the question,

                     Cu                                 S

Mass:            40.80 g                      51.09-40.80 = 10.29 g

Mole ratio:     40.80/63.5                 10.29/32.1

                         0.64          :                0.32

Divide by the smallest,

                         0.64/0.32   :                0.32/0.32

                           2                :                  1

   Therefore,

Empirical formula = Cu₂S.

Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-

Answers

Answer:

See explanation

Explanation:

A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.

Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.

The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.

According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.

Answers

Answer: 3. Molecules that are improperly oriented during collision will not react.

Explanation:

According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.

It depends on two factors:-

1. Energy factor:  For collision to be effective,  the colliding molecules must have energy more than a particular value called as threshold energy.

2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.

Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.

bonding is similar to ionic bonding, except there are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate. Group of answer choices

Answers

The question is incomplete, the complete question is;

["covalent", "Van der Waals", "ionic", "hydrogen", "metallic"] bonding is similar to ionic bonding, except their are no high-electronegativity atoms present to accept any electrons that the present atoms are willing to donate.

Answer:

metallic

Explanation:

The metallic bond bears a striking similarity to the ionic bonds only that there are no electronegative elements present to accept electrons in a metallic bond as in an ionic bond.

Most metals usually have a few valence electrons which are loosely bound to the outermost shell of the metal atom. Metallic bonds are usually comprised of metal ions bonded together by a sea of mobile electrons

These mobile electrons exert an attractive force on the positive ions and hold them together in the metallic crystal lattice. This force of attraction that holds the metal atoms together in the metallic crystal lattice is known as the metallic bond.

some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?

Answers

Answer:

3.97 L

Explanation:

Data obtained from the question include the following:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 19 °C

Final temperature (T2) = 58 °C

Final volume (V2) =..?

Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:

Temperature (K) = temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273 = 292 K

Final temperature (T2) = 58 °C

Final temperature (T2) = 58 °C + 273 = 331 K

Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 292 K

Final temperature (T2) = 331 K

Final volume (V2) =..?

V1 /T1 = V2 /T2

3.5 /292 = V2 /331

Cross multiply

292 x V2 = 3.5 x 331

Divide both side by 292

V2 = (3.5 x 331) / 292

V2 = 3.97 L

Therefore, the new volume of the gas is 3.97 L.

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated

Answers

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

Which of the following reagents should be used to convert to Question 26 options: A) H2, Lindlar's catalyst B) Na, liquid NH3 C) H2 / Nickel D) H2SO4, H2O

Answers

Answer: The question is is not complete...here is the complete question.

Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene

Option B.

Na, liquid NH3.

Explanation:

3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.

The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.

3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.

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