Which one of the following series of lines in the hydrogen spectrum arises from transitions down to n = 2?
A) 121, 102, 97, 95 nm
B) 655, 485, 433, 409 nm
C) 1872, 1279, 1092, 1003 nm
D) 4044, 2620, 2162, 1941 nm

Answer: mole fractions are

For n-pentane = 0.965

For n-octane = 0.035

Explanation: pressure exerted by each gas is,

n-pentane = 1.92atm

n-octane = 0.07atm

Total pressure exerted = 1.92 + 0.07

= 1.99atm.

Recall that the partial pressure exerted by each gas is the product of its mole fraction and the total pressure, that is,

Pres. n-pentane = n x pressure(total)

1.92 = n x 1.99

n = 1.92/1.99 = 0.965 for n-pentane

For n-octane,

n = 1 - 0.965 = 0.035 for n-octane.

Answer : The final concentration of ONGP is, 0.47 mM

Explanation :

Formula used :

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume

[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume

We are given:

[tex]M_1=3.3mM\\V_1=1.42mL\\M_2=?\\V_2=10mL[/tex]

Now put all the given values in above equation, we get:

[tex]3.3mM\times 1.42mL=M_2\times 10mL\\\\M_2=0.47mM[/tex]

Hence, the final concentration of ONGP is, 0.47 mM

Answer:

price ($) Au = $ 19183.94

Explanation:

april 15,2000:

∴ price Au = $ 282/t.oz

∴ 1.00 t.oz = 28.4 g

∴ V Au = 100.0 cm³  ⇒  price ($) = ?

∴ δ Au = 19.32 g/cm³

⇒ mass Au = (100.0 cm³)*(19.32 g/cm³)

⇒ m Au = 1932 g

⇒ price ($) = (1932 g Au)*(1.00 t.oz/28.4 g Au)*( $ 282/t.oz)

⇒ price ($) = $ 19183.94

To calculate the cost of 100.0cm3 of gold on April 15, 2000, convert the volume to grams, then to troy ounces, and multiply by the price per troy ounce.

To calculate the cost of 100.0cm3 of gold on April 15, 2000, we need to convert the volume of gold to grams and then to troy ounces. First, convert 100.0cm3 to grams by multiplying it by the density of gold (19.3 g/cm3). This gives us 1930 grams. Next, convert grams to troy ounces by dividing by the conversion factor of 28.4 grams per troy ounce. This gives us approximately 67.96 troy ounces. Finally, multiply the number of troy ounces by the price per troy ounce to find the cost. Therefore, 100.0cm3 of gold on April 15, 2000 would have cost $19,191.12.

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The distraction during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0. The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.

The distraction that Alicia experienced when the color change occurred during Kinetic Trial 2 will not affect the slope of the log (rate) vs log [I-]0.

This is because the distraction only affected the timing of the color change, not the actual reaction rate.

The relationship between the log of the rate and the log of the initial concentration of I- is determined by the reaction kinetics and is not influenced by external factors like distractions.

Answer: Please see answer below

Explanation:

Mecury vapor lamp is better to use than Sodium vapor light, this is because  because

---The Filaments of the lamp in sodium  emit fast moving electrons, which causes valence electrons of the sodium atoms to excite to higher energy levels which when electrons after being excited, relax by emitting yellow light which concentrates on the the  monochromatic bright yellow part of the visible spectrum which is about  580-590 or about (589nm) which will fall incident on the calibrations making it difficult to see

While

In Mercury vapor lamp, The emitted  electrons from the filaments, after having been excited  by high voltage, hit the mercury atoms but the excited electrons of mercury atoms relax and emits an  ultraviolet  uv invisible lights falling on the mecury vapour lamp to produce white light  covering  a wide range of  (380-780 nm) which is visible that is why it is used for calibrations purposes in  lightening applications.

Answer:

For N₂F₂:

Molar fraction = 0.84

Partial pressure = 1.12 atm

For SF₄:

Molar fraction = 0.16

Partial pressure = 0.208 atm

Explanation:

It seems your question is missing the values required to solve the problem. However, an internet search showed me the following values for your question. If the values in your problem are different, your answer will be different as well, however the solving method will remain the same:

" A 5.00L tank at 0.7°C is filled with 16.5g of dinitrogen difluoride gas and 5.00g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. "

First we calculate the moles of each gas, using their molar mass:

Total mol number = 0.25 + 0.0463 = 0.2963 mol

Now we use PV=nRT to calculate the partial pressure of each gas:

P = ?

V = 5.00 L

T = 0.7 °C ⇒ 0.7 + 273.16 = 273.86 K

For N₂F₂:

For SF₄:

Final answer:

To calculate the mole fraction, you divide the number of moles of a gas by the sum of the moles of all gases in the mixture. The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. The total pressure in the tank is the sum of the partial pressures of each gas.

Explanation:

The mole fraction of a gas is the ratio of the moles of that gas to the total moles of all gases in the mixture. To calculate the mole fraction of dinitrogen difluoride (N2F2), divide the moles of N2F2 by the sum of the moles of N2F2 and CO2:

Mole fraction of N2F2 = moles of N2F2 / (moles of N2F2 + moles of CO2)

To calculate the mole fraction of carbon dioxide (CO2), divide the moles of CO2 by the sum of the moles of N2F2 and CO2:

Mole fraction of CO2 = moles of CO2 / (moles of N2F2 + moles of CO2)

The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. To calculate the partial pressure of N2F2, multiply its mole fraction by the total pressure in the tank:

Partial pressure of N2F2 = mole fraction of N2F2 * total pressure

To calculate the partial pressure of CO2, multiply its mole fraction by the total pressure in the tank:

Partial pressure of CO2 = mole fraction of CO2 * total pressure

The total pressure in the tank is the sum of the partial pressures of N2F2 and CO2:

Total pressure = partial pressure of N2F2 + partial pressure of CO2

Answer:

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

The coefficient that will be used for Cl₂ in this reaction is 3

Explanation:

We use the method of electron-ion to the balance.

We assume that the redox reaction is happening at acidic medium.

Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺

Chloride is raising the oxidation state from -1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation

2Cl⁻ → Cl₂ + 2e⁻         Oxidation

In the dichromate anion, chromium acts with +6 in oxidation state, and we have 2 Cr, so the global charge of the element is +12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O   Reduction

As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.

(2Cl⁻ → Cl₂ + 2e⁻) ₓ3        

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O)  ₓ1    

We multiply the half reactions, in order to remove the electrons and we sum, the equations:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O

Now, that we have the same amount of electrons, they can cancelled, so the balanced redox reaction is:

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

To balance the given redox reaction, follow the steps: assign oxidation states, balance the elements undergoing oxidation or reduction, balance the elements not involved in redox, balance the charge, balance oxygen atoms, combine H2O molecules, and finally, balance the equation. The coefficient for Cl2 (g) is 3.

To balance the redox reaction Cr2O72- + Cl- → Cl2 + Cr3+ in an acidic solution, we need to follow these steps:

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Answer:

instantaneous rate at 40 s= 0.0035 M /s.

Explanation:

Instantaneous rate at 40 s is the slope of the line (tangent to the curve)

=  Δp/Δt

From, the straight orange line

ΔP = (0.48 - 0.16) M.

Δt = (92 -0) s

Now, instantaneous rate at 40 s

=  0.48 - 0.16/92 - 0

instantaneous rate at 40 s= 0.0035 M /s.

Answer:

0.0035

Explanation:

check the picture

Answer : The half-life of the reaction in seconds is, 244

Explanation :

The expression used for zero order reaction is:

[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]

where,

[tex]t_{1/2}[/tex] = half-life of the reaction = ?

[tex][A_o][/tex] = initial concentration = 5.90 M

k = rate constant = 0.725 M/min

Now put all the given values in the above formula, we get:

[tex]t_{1/2}=\frac{5.90}{2\times 0.725}[/tex]

[tex]t_{1/2}=4.069min=244.14s\approx 244s[/tex]

conversion used : (1 min = 60 s)

Thus, the half-life of the reaction in seconds is, 244

Answer:

10.1 g of FeCl₃ are formed by the reaction

Explanation:

First step is to determine the reaction where the reactants are Fe₂O₃ and HCl  in order to produce FeCl₃ and H₂O.

Equation is: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

We assume the acid is in excess, so the limiting reagent will be the oxide.

Let's work with mass:

1 mol of Fe₂O₃ is 159.7 g

2 mol of FeCl₃ is 162.2 g

So now we propose a rule of three:

159.7 g of oxide can produce 162.2 grams of chloride

Then, 10 g of oxide will produce (10 . 162.2) / 159.7 = 10.1 g of FeCl₃

Answer:

20.3 grams of FeCl3 will be formed

Explanation:

Step 1: Data given

iron(III) oxide = Fe2O3

hydrochloric acid = HCl

iron (III) chloride = FeCl3

water = H2O

Mass of Fe2O3 = 10.0 grams

Molar mass Fe2O3 = 159.69 g/mol

Step 2: The balanced equation

Fe2O3 +6HCl → 2FeCl3 + 3H2O

Step 3: Calculate moles Fe2O3

Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3

Moles Fe2O3 = 10.0 grams/ 159.69 g/mol

Moles Fe2O3 = 0.0626 moles

Step 4: Calculate moles of FeCl3

For 1 mol Fe2O3 we need 6 moles HCl to produce 2 moles FeCl3 and 3 moles H2O

For 0.0626 moles Fe2O3 we'll have 2*0.0626 = 0.1252 moles FeCl3

Step 5: Calculate mass FeCl3

Mass FeCl3 = moles FeCl3 * molar mass FeCl3

Mass FeCl3 = 0.1252 moles * 162.2 g/mol

Mass FeCl3 = 20.3 grams

20.3 grams of FeCl3 will be formed

Answer:

Molarity = 2.25M

Explanation:

n= m/M= 56/26=2.15mol, V= 959ml= 0.959L

n=C×V

2.15= C× 0.959

Simplify

C= 2.25M

Answer:

2.2M

Explanation:

Answer:

See detailed mechanism in the image attached

Explanation:

The mechanism shown in detail below is the synthesis of serine in steps.

The first step is the attack of the ethoxide ion base on the diethyl acetamidomalonate substrate giving the enolate and formaldehyde.

The second step is the protonation of the oxyanion from (1) above to form an alcohol as shown.

Acid hydrolysis of the alcohol formed in (3) above yields a tetrahedral intermediate, a dicarboxyamino alcohol.

Decarboxylation of this dicarboxyamino alcohol yields serine, the final product as shown in the image attached.

If all of the energy from the pancake is transferred into the water, the temperature of the water will increase by 0.02 °C

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

The increase in temperature can be obtained as follow:

Q = MCΔT

Divide both side by MC

ΔT = Q / MC

ΔT = 61 / (2500 × 1)

ΔT = 61 / 2500

ΔT = 0.02 °C

Thus, the temperature of the water will increase by 0.02 °C

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Answer:

The most appropriate structure given the sparse spectral data is 4-acetyl benzoic acid (see attached).

Explanation:

It is difficult to accurately elucidate the structure of this compound without its chemical formula. But from the 1H NMR spectral data shows a total of 8 hydrogen atoms:

The attached files show the structure and the neighboring hydrogen atoms.

The most likely structure i 4-acetyl benzoic acid

Answer:

Explanation:

During titration indicators are often used to identify chemical changes between reacting species.

For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.

Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.

Most of these indicators have various colors when chemical changes occur.

Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.

Final answer:

An observation indicative of a chemical change during titration could include temperature change, light emission, unexpected color change, or the formation of bubbles signaling the production of gas, any of which suggest that a new substance has been formed.

Explanation:

For instance, if the student added an indicator to the acid and after titrating with a base, noticed a color change, this would be evidence of a chemical change. Similarly, if the student observed the solution fizzing but it was not reaching boiling temperature, it could indicate the formation of a gaseous product.

Answer : The good buffer systems are, (a), (b), (c) and (e)

Explanation :

Buffer : It is a solution that prevent any changes in the pH of the solution on the addition of an acidic and basic components.

Or, it is a solution that maintain the pH of the solution by adding the small amount of acid or a base.

There are two types of buffer which are acidic buffer and basic buffer.

Acidic buffer : It is the solution that have the pH less than 7 and it contains weak acid and its salt. For example : Acetic acid (weak acid) and sodium acetate (salt).

Basic buffer : It is the solution that have the pH more than 7 and it contains weak base and its salt. For example : Ammonia (weak base) and ammonium chloride (salt).

The conditions for a good buffer system is:

(1) a weak acid and its conjugate base.

(2) a weak base and its conjugate acid.

(a) 0.37 M hydrocyanic acid + 0.24 M sodium cyanide

It is a combination of weak acid and its conjugate base. So, it is a good buffer.

(b) 0.14 M sodium nitrite + 0.27 M nitrous acid

It is a combination of weak acid and its conjugate base. So, it is a good buffer.

(c) 0.21 M hypochlorous acid + 0.13 M potassium hypochlorite

It is a combination of weak acid and its conjugate base. So, it is a good buffer.

(d) 0.20 M nitric acid + 0.22 M potassium nitrate

It is a combination of strong acid. So, it will not form buffer solution.

(e) 0.32 M ammonium bromide + 0.31 M ammonia

It is a combination of weak base and its conjugate acid. So, it is a good buffer.

Hence, the good buffer systems are, (a), (b), (c) and (e)

Good buffer systems are those that consist of a weak acid and its conjugate base or a weak base and its conjugate acid. The given combinations that are good buffer systems include hydrocyanic acid and sodium cyanide, sodium nitrite and nitrous acid, hypochlorous acid and potassium hypochlorite, and ammonium bromide and ammonia.

The aqueous solutions that are good buffer systems from the provided options are:

A good buffer system consists of a weak acid and its conjugate base or a weak base and its conjugate acid. These pairs are capable of neutralizing small amounts of added acid or base, thus maintaining a relatively stable pH in the solution. In contrast, solutions like 0.20 M nitric acid + 0.22 M potassium nitrate do not form a buffer because nitric acid is a strong acid and does not create a conjugate weak acid-base pair needed for buffering.

Answer:

The factor is 2

Explanation:

Van't Hoff factor is defined as the ratio between the species of a solute before the addition to the solvent and particles produced when the substance is dissolved. It is used, principally, in colligative properties.

Before solution, potassium bromide, KBr, has just one specie, that is, KBr. When KBr is dissolved (As a salt):

KBr(aq) → K⁺(aq) + Br⁻(aq)

There are produced two species, K⁺ and Br⁻. By definition of Van't Hoff factor, for this salt, the factor is 2.

Answer:

The chemical equation is given as:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

Explanation:

When gaseous ammonia reacts with gaseous oxygen it gives nitrogen monoxide gas and water vapors as product.

The chemical equation is given as:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

According to reaction, 4 moles of ammonia reacts with 5 moles of oxygen gas to give 4 moles of nitrogen monoxide gas and 6 moles of water vapor.

To write the skeletal equation, begin by writing the chemical formula for each reactant and product.

Reactants:

1. Ammonia's chemical formula is given as  NH3.

2. Gaseous oxygen exists as a diatomic molecule with the chemical formula  O2.

Products:

1. Nitrogen monoxide has the chemical formula of  NO.

2. The chemical formula for water is  H2O.

Therefore, the skeletal equation is written as:

NH3(g)+O2(g)→NO(g)+H2O(g)

Now count the number of each atom on each side of the equation to determine if the equation is balanced.

Reactants

1N atom

3H atoms

2O atoms

Products

1N atom

2H atoms

2O atoms

Begin by balancing the number of hydrogen atoms by adding a coefficient of 2 to  NH3 and a coefficient of 3 to  H2O. Next, balance the number of nitrogen atoms by adding a coefficient of 2 to  NO. Now there are two oxygen atoms on the reactant's side and five oxygen atoms on the product's side of the reaction. Since only whole number coefficients should be used, all coefficients need to be increased by a factor of two to balance the oxygen atoms. Thus the coefficient for  NH3 is 4, the coefficient for  H2O is 6, and the coefficient for  NO is 4. Finally, balance the oxygen atoms by adding a coefficient of 5 to  O2. The balanced equation is:

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

Answer:

The weakest oxidizing agent is Zn^2+(aq)

The strongest reducing agent is Zn(s)

The strongest oxidizing agent is I2(s)

The weakest reducing agent is I^-(aq)

I^- cannot reduce Zn^2+ to Zn(s)

I2(s) can be reduced by hydrogen gas

Explanation:

In looking at oxidizing and reducing agents, our primary guide is the reduction potentials of each specie. The more negative the reduction potential of a specie, the better its function as a reducing agent. Zn has a very negative reduction potential hence it a very good reducing agent. Similarly, iodine has a very positive reduction potential hence it is a good oxidizing agent.

Only a specie having a more negative reduction potential than zinc can reduce it in aqueous solution. Similarly, the reaction potential of hydrogen is less than that of iodine hence hydrogen gas can reduce iodine.

Final answer:

In electrochemistry, the strongest reducing agent is Zn(s) and the weakest oxidizing agent is Zn2+(aq). I2(s) is the strongest oxidizing agent, while I-(aq) is the weakest reducing agent. I-(aq) cannot reduce Zn2+(aq) to Zn(s), and none of the species provided can be reduced by H2(g).

Explanation:

To determine the oxidizing and reducing agents, we refer to their standard electrode potentials (E°). A strong oxidizing agent has a higher positive E° value, indicating a greater tendency to gain electrons and be reduced. Conversely, a strong reducing agent has a lower (more negative) E° value, reflecting a higher tendency to lose electrons and be oxidized.

The weakest oxidizing agent is Zn²+(aq) because it has the most negative E° value (-0.763V), meaning it is the least likely to gain electrons. The strongest reducing agent is Zn(s) because zinc in its solid state is more willing to be oxidized (lose electrons) as indicated by its half-reaction (Zn(s) → Zn²+(aq) + 2e⁻).

Similarly, the strongest oxidizing agent is I₂(s) due to its higher E° value (0.535V), which demonstrates its greater ability to take up electrons. The weakest reducing agent in this set is I⁻(aq) because it is derived from I₂(s), which is a strong oxidizer, thus its conjugate, I⁻, would be the weakest reducer.

Regarding whether I⁻(aq) can reduce Zn²+(aq) to Zn(s), the answer is no, because I⁻(aq) is the weakest reducing agent and Zn²+(aq) is not a strong oxidizer. Lastly, H₂(g) has an E° value of 0.000V, making it neutral in this context, and therefore it will not reduce any of the species provided in the question.

Answer:6.0 ML

Explanation:

Answer:

Explanation:

the solution is solved below

When ethers react with HI, treatment with one equivalent of HI produces an alcohol and an alkyl iodide as major products. Treatment with excess HI yields both alkyl iodides as major products.

Ethers react with HI to form two cleavage products. When treated with one equivalent of HI, the major products that could be recovered are an alcohol and an alkyl iodide. The alcohol is formed by the substitution of the ether oxygen with a hydrogen atom from HI, and the alkyl iodide is formed by the substitution of one of the alkyl groups of the ether with iodine.

When treated with excess HI, the major products that could be recovered are both alkyl iodides. The initial products from the first step do not further react but are still recovered.

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Answer: 2Mg (s) + O2 (g) ----> 2MgO (s)

Explanation:

Lets write the equation of the reaction.

Mg(s) + O2 (g) ----> MgO (s)

Counting the number of atoms for each element we have:

Left hand side: Mg =1 O = 2

Right hand side : Mg = 1, O =2

To balance this equation input "2" as coefficient for "MgO" on the right hand side and "2", as coefficient for "Mg" on the left hand side of the equation. Hence our balanced equation will be

2Mg (s) + O2 (g) -----> 2MgO (s)

Final answer:

The balanced chemical equation for the combination reaction between magnesium metal and gaseous oxygen to form magnesium oxide is: 2Mg (s) + O₂(g) → 2MgO (s). This reaction follows the conservation of mass.

Explanation:

The reaction between magnesium metal and gaseous oxygen is a combination reaction in which magnesium is oxidized. When magnesium (Mg) combines with oxygen (O₂), it forms magnesium oxide (MgO). The equation representing this exothermic reaction is written as:

2Mg (s) + O₂(g) → 2MgO (s)

This reaction adheres to the law of conservation of mass, meaning the total mass of the reactants equals the total mass of the products. During the reaction, magnesium atoms lose electrons (are oxidized) and the oxygen molecule gains electrons.

Ideal gas law is valid only for ideal gas not for vanderwaal gas. The equation used for ideal gas is PV=nRT. The  number of molecule-wall collisions per unit area per unit time will remain the same.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically the relation between Pressure, volume and temperature can be given as

PV=nRT

where,

P = pressure of gas

V= volume of gas

n =number of moles of gas

T =temperature of gas

R = Gas constant = 0.0821 L.atm/K.mol

If the volume of the gas sample is decreased to 15.7 L holding the temperature constant, the number of molecule-wall collisions per unit area per unit time will remain the same.

Therefore, the number of molecule-wall collisions per unit area per unit time will remain the same.

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The Question is incomplete here is the complete question " Suppose an iron atom in the oxidation state +3 formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.

Answer:

{Fe(OH)3(H20)3}

Explanation:

Oxidation state is the electron gained or lost by and atom, So if iron is in +3 state in formula it must have lost three electron.

We know that OH posses the oxidation state of -1 and water have zero oxidation state. SO, Let's take iron equal to y and find its oxidation state in the formula

y + 3 ( - 1 ) + 3 ( 0 ) = 0

y - 3 + 0 = 0

y-3=0

y= + 3

Hence it's proved that iron has +3 oxygen state.

Answer : The correct option is, (D) [tex]9.0\times 10^9J[/tex]

Explanation :

As we are given that the energy require for decomposition is, [tex]9.0\times 10^6kJ[/tex].

Now we have to calculate the energy in joules.

Conversion used :

1 kJ = 1000 J

As, 1 kJ of energy = 1000 J

So, [tex]9.0\times 10^6kJ[/tex] of energy = [tex]\frac{9.0\times 10^6kJ}{1kJ}\times 1000J[/tex]

                                          = [tex]9.0\times 10^9J[/tex]

Therefore, the energy in joules is, [tex]9.0\times 10^9J[/tex]

Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 × 10⁹ Joules of heat.

Enegy is the quantitative property which is used by any system to perform any work.

Chemical reactions generally involves energy in the form of heat energy and given amount of energy is 9.0 × 10⁶ kJ.

We know that:

1 kJ = 1000 J

So, 9.0 × 10⁶ kJ = 9.0 × 10⁶ kJ  × 1000

9.0 × 10⁶ kJ = 9.0 × 10⁹ J

Hence, option (D) is correct i.e. 9.0 × 10⁹ J.

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Answer:

Aldol condensation is possible only when their is alpha Hydrogen atom is present. It ia present only in the acetophenone and not in benzaldehyde.

Explanation:

Answer:

57.5%

Explanation:

The mass percent of copper in malachite (Cu2(OH)2CO3) can be determined as follow:

Molar Mass of malachite (Cu2(OH)2CO3) = (2x63.5) + 2(16 +1) + 12 + (16x3) = 127 + 2(17) + 12 + 48 = 127 + 34 + 12 + 48 = 221g/mol

Mass of Cu in Cu2(OH)2CO3 = 2 x 63.5 = 127g

The percentage by mass of Cu in Cu2(OH)2CO3 is given by:

Mass of Cu/Molar Mass of Cu2(OH)2CO3 x 100

=> 127/221 x 100

=> 57.5%

Therefore, 57.5% by mass of Cu is contained in malachite Cu2(OH)2CO3

Answer:

Solution is 4.67% by mass of salt

Explanation:

% by mass is the concentration that defines the mass of solute in 100g of solution.

In this case we have to find out the mass of solution with the data given:

Mass of solution = Mass of solute + Mass of solvent

Solute:  Salt → 14.2 g

Solvent: Water → 290 g

Solution's mass = 14.2 g + 290g = 304.2 g

% by mass = (mass of solute / mass of solution) . 100

(14.2 g / 304.2g) . 100 = 4.67 %