Which statement best describes inorganic compound?
A. Inorganic compounds are compounds do that not contain carbon and hydrogen.
B. Inorganic compounds are compounds that do not contain carbon atoms only.
C. Inorganic compounds are compounds that produced by living things.
D. Inorganic compounds are compounds that contain carbon and oxygen only.

To calculate the volume of ethyl alcohol that evaporated, we need to know the volume of ethyl alcohol that was present initially, the volume of ethyl alcohol after the evaporation, and the molar mass of ethyl alcohol.

Let's assume that the initial volume of ethyl alcohol is V_0 mL and the final volume is V_f mL. The volume of ethyl alcohol that evaporated is then V_e = V_0 - V_f mL.

We can use the ideal gas law to relate the volume of a gas to the number of moles of the gas present. The ideal gas law is given by:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Since the volume of a gas is directly proportional to the number of moles of the gas, we can write:

V/n = k

where k is a proportionality constant.

If we know the value of k for a given gas at a particular temperature and pressure, we can use this relationship to calculate the number of moles of the gas present.

In this case, we are told that 1 mole of any gas at STP (standard temperature and pressure) occupies a volume of 22.4 L, which is equivalent to 22,400 mL. Therefore, the value of k for any gas at STP is 22,400 mL/mol.

We can use this value of k to calculate the number of moles of ethyl alcohol that evaporated:

n = V_e/(22,400 mL/mol)

Substituting in the known values for V_e and k, we can calculate the number of moles of ethyl alcohol that evaporated.

For example, if the initial volume of ethyl alcohol was 1,000 mL and the final volume was 900 mL, the volume of ethyl alcohol that evaporated would be V_e = 1,000 mL - 900 mL = 100 mL. The number of moles of ethyl alcohol that evaporated would be:

n = 100 mL/(22,400 mL/mol) = 0.0044 moles

I hope this helps! Let me know if you have any questions.

if dfb = 3 and dft = 29 and fobt = 3.15, using the given we can can reject our null hypothesis, H_0.

Option (a) is the correct choice.

From the available information,

d fB=3, d fT=29, alpha=0.05, F_{obt}=3.15

Then, dfw = dfT-dfB = 29-3 = 26

In variance analysis, the F value is utilized (ANOVA). Two mean squares are divided to calculate it.

The ratio of explained variance to unexplained variance is calculated using this formula. A theoretical distribution is the F distribution.

In hypothesis testing, F Statistics are looked up using the F Table. Although it's more typical to perform tests using software like Excel or SPSS, the F Table might be helpful for rapidly searching up several data at once.

From the F-table,

The tabulated value at F_{0.05}(3,26) is 2.975.

Here the calculated value 3.15 is greater than the tabulated value 2.975,

So, we can reject our null hypothesis, H_0.

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The complete question should be like:

if dfb = 3 and dft = 29 and fobt = 3.15, what would you conclude using a = 0.05?

(a) reject  H_0

(b) reject  H_1

(c) retain  H_0

(d) retain  H_1

Answer:

Yeah. Mhm. So on this question, it asked about finding the highest density for a few gasses all at the same temperature. And since gasses will all have the same volume, if they have the same number of moles, your density would be The mass of each over the volume of 25°C. Now the volume of 0°C is 22.4. In order to find the volume at 25°C, you could do PV Equals NRT and plug in one mole, 25°C Plus 2 73. Right to get to Calvin's. Um The pressure value which I don't believe was given in the question of one atmosphere of pressure. Okay, so one atmosphere of pressure, the R value and atmospheres and then you would find the volume. Each of them. You would take their molar mass over that volume. Or if all you want to do is rank them. The highest molar mass would be the most dense. So if you needed the actual density, okay, you would do the molar mass Over the volume. For one mole, each gas would have the same volume. But if all you have to do is rank them. The one with the highest molar mass is the most ends and the one with the lowest molar mass would be the least ends.

Explanation:

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If it starts off at 64 grams, there are just [tex]\frac{1}{4}[/tex] grams left after 9 days.

The half-life of a radioactive element is the amount of time it takes for it to decay to half of its original value. This means that a source's activity has a half-life when it takes a certain amount of time for it to decrease to half of what it was. These radioactive atoms release energy at a quantifiable pace known as radioactive decay to transform into new, distinct sorts of atoms.

[tex]r= \frac{1}{2}\\[/tex]

[tex]a_{1} = 64\\n=9[/tex]

[tex]a_{n}= a_{1}(r)^{n-1}[/tex]

[tex]a_{9}= 64(\frac{1}{2} )^{9-1}\\a_{9}= 64(\frac{1}{2} )^{8}\\a_{9}=64(\frac{1}{256} )\\a_{9} = \frac{64}{256} \\a_{9}= \frac{1}{4}[/tex]

[tex]a_{9} = \frac{1}{4}[/tex] grams

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Ag (aq) → Ag⁺ (aq) + 1 e⁻ at the anode and NO₃ (aq) + 1e⁻ → NO₃⁻ at the cathode.

The half cell reaction are as follows:

In the anode oxidation always occurs, during the oxidation the reducing agent loses electrons, and its oxidation number increases.

NO₃ (aq) + 1e⁻ → NO₃⁻

In the cathode reduction always occurs, during the reduction the oxidizing agent gains electrons, and its oxidation number decreases.

Ag (aq) → Ag⁺ (aq) + 1 e⁻

Hence, Ag (aq) → Ag⁺ (aq) + 1 e⁻ occurs at the anode and NO₃ (aq) + 1e⁻ → NO₃⁻ occurs at the cathode.

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The Ground state electronic configuration of  Cr2+:[Ar] 3d4, Cu2+: [Ar] 3d9, Co3+: [Ar] 3d6

Electronic configuration, sometimes referred to as electronic structure or electron configuration, is the arrangement of electrons in orbitals that surround an atomic nucleus.

1. Cr2+

Atomic no.of Chromium(Cr)=24  

a ground state Chromium's electronic structure is as follows: [Ar] 4s1 3d5 = 1s2 2s2 2p6 3s2 3p6

(Atomic no.of [Ar](argon) = 18)

so,ground state electronic configuration for Cr2+:

[Ar] 4s0 3d4 =[Ar] 3d4

so, ground state electronic configuration of Cr2+:

[Ar] 3d4

2. Cu2+

Atomic number of Copper(Cu)= 29

Ground state electronic configuration of Cu:

[Ar] 4s1 3d10

so, ground state electronic configuration of Cu2+:

[Ar] 4s0 3d9 = [Ar] 3d9

so,electronic configuration of Cu2+:

[Ar] 3d9

3.Co3+

Atomic number of Cobalt(Co)=27

Ground state electronic configuration of Co:

[Ar] 4s2 3d7

so,electronic configuration of Co3+:

[Ar] 4s0 3d6 = [Ar] 3d6

So,electronic configuration of Co3+:

[Ar] 3d6

Therefore the Ground state electronic configuration of

Cr2+:[Ar] 3d4, Cu2+: [Ar] 3d9, Co3+: [Ar] 3d6

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si and ar pair of atoms have the most similar atomic radius

The distance between identical atoms' bound nuclei is known as the atomic radius, and it is equal to one-half of that distance. Illustration 6.15. 2 One-half of the distance (d) between the two nuclei of a diatomic molecule is known as the atomic radius, or r, of an atom. The diameter of an atom is determined by the space between its repelling atoms' nuclei; the radius is equal to half of this distance. Consider an illustration. 198 picometers (pm), or 1.98, is the diameter created by the collision of chloride (Cl) atoms (angstroms). (An angstrom () is equal to 10-10 metres.

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The stable isotopes have an infinite half-life. It's hard to get longer than infinite. There are 252 stable isotopes.

The longest known half-life is currently tellurium-128 at 7.7 × 10^24 years.

The longest directly measured half-life is presently xenon-124 with 1.8 × 10^22 years.

It is somehow possible that some isotopes that we currently think are stable might turn out to have a very long half-life, but we haven't figured it out yet.

For example, scientists used to think that bismuth-209 was stable.

In 2003 we discovered it is radioactive with a half-life of 2.01 × 10^19 years.

Stable isotopes are “stable" and therefore not radioactive. An atomic nucleus needs to be “unstable” for it to be radioactive.

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Only E alkene is produced as a significant product in this reaction. Z-alkenes exhibit more steric repulsions because the bulky isopropyl and methyl groups are both present on the same side of the double bond.

The study of the relative spatial arrangement of the atoms that are present in the molecules' structure is known as the stereochemistry of the substance. Isomerization occurs when two molecules have the same chemical structure but distinct configurations.

The stereoisomers are the cis-trans isomers. The two comparable or higher priority groups are linked to the carbon atom with the double bond on the same side in the cis isomer. The two comparable or higher priority groups are connected to the carbon atom with the double are on the opposite side in the trans isomer.

Saytzeff's rule states that when an asymmetric molecule is eliminated and there are many places where the unsaturation might occur, the alkene with the greatest number of substituted carbons will often result.

When a base and 3-Bromo-2,2,3-trimethyl pentane react, two compounds are produced by dehydrohalogenation. One product has more substituted alkene, whereas the other has less.

The reaction is shown in the below-mentioned diagram.

Saytzeff's Rule states that alkenes with more substituents are more stable.

As a result, the key component of the given reaction is illustrated below:

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Sn is more metallic than Te because following the path between Sn and Te in the periodic table moves to the right within the same period.

The metallic properties of an element are defined as the susceptibility of its atoms to lose electrons. According to the modern periodic table, the metallic properties of elements decrease as they move from left to right over a period of time. Moving cyclically increases the nuclear attraction due to the increase in atomic number and decreases the atomic size. Therefore, elements cannot readily donate electrons. Therefore, metallicity decreases over a period moving from left to right. Moving down the group increases the metal letters on the periodic table. As the distance between the valence electrons and the nucleus increases, the attractive force between them also decreases. If you increase the atomic radius, you lose electrons very easily.

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The heat of solution when solid sodium hydroxide dissolve in water is: eq Delta H =-44.4, 37.76 C is the final temperature of a solution.

Given data:

∆H = -44.4 kJ/mol

The heat released by 1 mol of NaOH is:

1 mol NaOH×(44.4 kJ)/(1 mol NaOH)×(1000 J)/(1 kJ)=44400 J

The heat released by 13.9 g of NaOH solution is:

13.9 g NaOH×(1 mol NaOH)/(40 g NaOH)×(44400 J)/(1 mol NaOH)=15429 J

The temperature is:

Q=mC(T_2-23)

15429=250×4.18×(T-23)

Solving for T final temperature

T final temperature = 37.76°C

The amount of heat released or absorbed during a reaction in solution that takes place under constant pressure is referred to as the enthalpy change. It has the sign H, which may be interpreted as "delta H."

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2-4cm of magnesium ribbon was cut.

The synthesis of cyclopentyl ethyl ether, a combination of bromocyclopentane and sodium ethoxide (option a) or ethanol and bromocyclopentane (option c) can be used as reaction conditions.

Synthesis is the process of combining multiple sources of information to create a unique idea or perspective. It involves taking a variety of ideas and combining them together to form a new, unique thought or opinion. Synthesis is often used in research papers and essays to develop a deeper understanding of a given topic. It allows readers to draw their own conclusions by looking at the various sources of information and forming their own opinion. Synthesis is an important skill to have in order to be able to think critically, analyze information, and form new ideas.

The Williamson Ether Synthesis (WES) is a method for synthesizing ethers from alcohols and alkyl halides. The reaction involves deprotonation of the alcohol with a base such as sodium ethoxide to form an alkoxide, followed by reaction with the alkyl halide.

For the synthesis of cyclopentyl ethyl ether, a combination of bromocyclopentane and sodium ethoxide (option a) or ethanol and bromocyclopentane (option c) can be used as reaction conditions. Therefore, the correct answer is (d) both a and c.

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(2R, 3R, 4S) - 2,3,4 - trichloro pentane and (2R,3R,4R)- 2,3,4- trichloro heptane both pairs are Enantiomers which exactly have the same connectivity byt opposite three- dimensional space.

The Enantiomers are non-superimposable mirror images of each other. In a mirror view absolute configuration R becomes S and S becomes R. This can be observed in the names of the two given isomers. Also, all other positions and basic skeleton structure of the compound is same in both the isomers. Two compounds with the exact same connectivity, that are mirror images of each other but that are not identical to each other. Enantiomers contain no mirror planes. Enantiomers do not contain two equal and opposite halves. Enantiomers are another example of a type of stereoisomers. Two enantiomers have identical physical properties, except for optical rotation.

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The freezing point of 0.5 mole of NaCl in 1.0 kg water will be -1.86 °C

The freezing point depression formula is:

Δ[tex]T = i * Kf * m[/tex]

where:

ΔT = the change in the freezing point

i = the van't Hoff factor of NaCl = 2

Kf = the freezing point depression constant of water = [tex]1.86\frac{Celcius}{m}[/tex]

m = the molality of the solution

The molality of the solution will be stated in the equation as moles of NaCl per kilogram of solvent.

Δ[tex]T = i * Kf * m[/tex]

Δ[tex]T = i * Kf * (\frac{moles NaCl}{kg of water} )[/tex]

We can now use the equation for the freezing point depression to calculate the freezing point of the solution.

Δ[tex]T = i * Kf * (\frac{moles NaCl}{kg of water} )[/tex]

Δ[tex]T = 2 * 1.86 * (\frac{0.5}{1} )[/tex]

Δ[tex]T = 1.86[/tex]

[tex]0.00 Celcius - Freezing Point Solution = 1.86[/tex]

[tex]Freezing Point Solution = -1.86 Celcius degree[/tex]

As the result, the freezing point of the solution will be -1.86 °C and the right answer is E.

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The rate orders would be: C. A = 2, B=0, C=1

The rate law demonstrates how the concentration of the reactant affects the rate of a chemical reaction. The rate law typically takes the form rate = k[A]n for a reaction such as a products, where k is a proportionality constant known as the rate constant and n is the order of the reaction with respect to A.

Option (1), (2) and Option 4 is incorrect because the provided expression is incorrect for a reaction rate.

The reaction is 2A + B + 2C > E. The mathematical dependence of the rate on reactant concentrations is described by the reaction orders in a rate law. The reaction is m order with respect to A and n order with respect to B, according to the general rate law discussed above. For instance, the reaction is first order in A and second order in B if m = 1 and n = 2.

In conclusion, the correct option is C.

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In the given reaction: 2Bi³⁺ + 3Mg → 2Bi + 3Mg²⁺ , Bi³⁺ is reduced to Bi.

In a chemical reaction, reduction means gain of electrons or loss of oxygen whereas oxidation means loss of electrons or gain of oxygen. An oxidizing agent is always reduced and a reducing agent is always oxidized.

Reactions in which oxidation and reduction takes place simultaneously are known as redox reactions. The above reaction is an example of redox reaction.

Magnesium (Mg ) being the reducing agent, reduces  Bi³⁺ to Bi (Bismuth) and is itself oxidized to Mg²⁺.

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The 7 moles of the sample remain after 45 minutes.

What is half-life?

The amount of time it takes for a particular reactant's concentration to reach 50% of its initial concentration is known as the half-life of a chemical reaction.

What is first order reaction ?

A first-order reaction is a chemical process in which the rate of the reaction depends linearly on the concentration of just one ingredient. To put it another way, a first-order reaction is a chemical reaction in which the rate varies depending on changes in the concentration of just one of the reactants.

Therefore, 7 moles of the sample remain after 45 minutes.

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Answer: 12.5 mol of unknown

Explanation: 15 minutes per generation after 45 minutes there will be 3 generations [tex]\frac{45}{15}[/tex] = 3.

1.0*10^2 = 100mol

100mol → 50mol → 25mol → 12.5mol or 13mol

Initial     1st gen   2nd gen    3rd gen

The weight/volume % concentration of the solution is 20% that is calculated by using the formula below.

Weight of the solute (glucose)= 5.00 g

Volume of solution in liter=0.025 L

Volume of solution in mL=25 mL

In order to calculate % wight/volume, the formula is used which is expressed as follows:

% weight/volume= mass of solute (in gm)/volume of solution (in mL) * 100

It is a measure of the concentration of a solution. w/v% or m/v% is calculated by dividing the mass of the solute in gm by the volume of solution in mL then multiplying this by 100.

Thus by substituting the values,

%weight /volume = 5.00 gm/25 mL*100

=20%

Thus, %weight/volume is 20%.

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The balanced chemical equation for the given reaction is given as,

2AgNO₃(aq) + NiCl₂(aq) → 2AgCl(s) + Ni(NO₃)₂(aq)

For writing a balanced chemical equation for the reaction between aqueous silver nitrate and aqueous nickel(II) chloride to produce solid silver chloride and aqueous nickel(II) nitrate

First, we have to write the chemical formula for the each of the reactants and products is,

Silver nitrate - AgNO₃

Nickel(II) chloride - NiCl₂

Silver chloride - AgCl

Nickel(II) nitrate - Ni(NO₃)₂

The balanced chemical equation for the reaction is given as,

2AgNO₃(aq) + NiCl₂(aq) → 2AgCl(s) + Ni(NO₃)₂(aq)

Hence, the balanced chemical equation for the reaction is

2AgNO₃(aq) + NiCl₂(aq) → 2AgCl(s) + Ni(NO₃)₂(aq)

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The mass of this isotope this is required to treatment of cancer with half life of 2.70 days and 205 C is 1.187 mg.

Isotope is described as or greater extraordinary forms of atoms with the identical atomic variety, however extraordinary nucleon values at extraordinary positions at the periodic table. Isotopes are varieties of an detail which have the identical variety of protons and electrons however differing numbers of neutrons.

Let's first translate the 2.70 day 1/2 of existence into seconds to achieve the subsequent 1/2 of existence: T(1/2) =233300 sec2.971x10-6, decide the decay constant.

Now that we've the decay constant, we are able to clear up for the amount of isotopes required to supply an pastime of 290 Ci (that's identical to 1.073x10¹³ decays consistent with second) the use of the fee of degradation equation. Due to the truth that we're discussing the start action, t = zero for this equation.

- 1.073 * 10 ^ 3 =-A?e^ ^ (-?*zero) Since we're dropping isotopes as opposed to obtaining them, the left facet of the equation is negative.

Fill withinside the blanks: A = 3.611x1018 isotopes

Find out how plenty mass is contained in all of those isotopes now.(3.611x1018) x(198u) = 7.150x102° u

Thus, the mass of this isotope this is required to present an pastime of 205 C is 1.187mg.

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The complete question

One of the many isotopes used in cancer treatment is 198Au, with a half-life of 2.70 days. Determine the mass of this isotope that is required to give an activity of 221 Ci.

Answer:The mass of aluminum sulfate required to produce 455 mL of 0.310 mol/L of aluminum ions is approximately 94.7 grams.

Explanation: To determine the mass of aluminum sulfate that is required to produce 455 mL of 0.310 mol/L of aluminum ions, you will need to first calculate the number of moles of aluminum ions that are required. You can use the following steps:

Convert the volume of the solution to liters. Since there are 1000 mL in 1 L, the volume in liters is:

volume (L) = 455 mL / 1000 mL/L = 0.455 L

Calculate the number of moles of aluminum ions. Using the concentration of the solution in mol/L, the number of moles of aluminum ions is:

moles (mol) = concentration (mol/L) x volume (L)

= 0.310 mol/L x 0.455 L = 0.139 mol

Calculate the mass of aluminum sulfate required to produce this many moles of aluminum ions. Aluminum sulfate contains two aluminum ions per molecule, so the number of moles of aluminum sulfate required is twice the number of moles of aluminum ions. The mass of aluminum sulfate required can then be calculated using the molar mass of aluminum sulfate:

mass (g) = moles (mol) x molar mass (g/mol)

= (0.139 mol x 342.1 g/mol) / 2

= 94.7 g

Therefore, the mass of aluminum sulfate required to produce 455 mL of 0.310 mol/L of aluminum ions is approximately 94.7 grams.

Keep in mind that this is the mass of aluminum sulfate required to produce the desired concentration of aluminum ions in the solution, and that the actual mass of the solution may be larger due to the presence of water or other dissolved substances.

2.8 x 10^9 atoms carbon atoms are there in a 2.6-carat diamond. diamonds are a form of pure carbon

Having the chemical symbol C and atomic number 6, carbon is an element with the meaning "coal" in Latin. It is nonmetallic and tetravalent, meaning that four of its atom's electrons can be used to create covalent chemical connections. The periodic table's group 14 is where it belongs. Only 0.025 percent of the crust of the Earth is made up of carbon. Crude oil, methane gas, and coal are used as fuel sources (which is used to make gasoline). All kinds of materials i.e diamond, including polymers and steel alloys, are made with it (a combination of carbon and iron)

2.8 carat C x (0.20g C/ 1 carat C) x (1 mole C/ 12.01 g C) x (6.02 x 10^23 atoms C/ 1 mole C)

= 2.8 x 10^9 atoms C

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Balanced Equation: CaSiO3 + 2 CO2 + H2O = Ca(HCO3)2 + SiO2

One or more chemical substances can react with each other and form one or more new substances through a process which is referred to as chemical reaction.

Reactants are the elements required for a reaction's initialization.

The components that remain after a reaction are called products.

Each kind of atom is to be present on both the reactant and product side of a chemical equation in a chemical reaction in order for it to be balanced.

On both sides of balanced chemical equations, the same number and type of each atom can be found.

A balanced chemical equation must have coefficients that are the smallest whole number ratio.

Mass is always conserved in chemical processes.

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In cranberry glass, an additive used to impact red coloration is Purple of Cassius.

The purple of Cassius is a pigment which can be prepared as follows;

When gold chloride reacts with tin(II) chloride or stannous chloride, a purple pigment is formed. This pigment is called Purple of Cassius.

If gold chloride and tin chloride are present in the solution, cranberry glass is formed.

The intensity of the red color depends upon the concentration of gold present.

This method is used to test the concentration of gold.

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DKA results from a lack of blood sugar entry into your cells, which occurs when your body doesn't create enough insulin to meet your needs. Instead, a procedure known as lipolysis occurs in your liver when fat is changed into acids that produce ketone bodies lead to the development of ketoacidosis.

DKA is less frequent and less severe in patients with type 2 diabetes. It is frequently caused by protracted periods of uncontrolled blood sugar, forgotten medication doses, or a significant illness or infection.

There are several types of ketoacidoses that are clinically significant, including diabetic ketoacidosis (DKA), alcoholic ketoacidosis (AKA), and starvation ketoacidosis. DKA is an untreated diabetic complication that could be lethal.

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Nuclear reactions are much more energetic than chemical reactions.

Nuclear reactions involve protons and neutrons in the nucleus. Chemical reactions involve electrons orbiting the nucleus. In an exothermic reaction, more energy is released than is used to break the bond of the reactants forming a bond in the product. Nuclear energy is millions of times more powerful than chemical energy.

An exothermic reaction involves an increase in the temperature of the reaction mixture. A chemical reaction that totally absorbs energy is called an endotherm. All chemical reactions require energy. However, like rust and combustion, not all chemical reactions release energy. Some chemical reactions absorb energy rather than release it.

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Answer: d)

Explanation: Sulfuric acid on reaction with sugar leads to the formation of a lot of heat indicating it to be an exothermic reaction, a large quantity of steam , a black carbon containing tube like structure pushing itself out of the beaker.It is demonstrated as:

C12H22O11 (sugar) + H2SO4 (sulfuric acid) → 12 C (carbon) + 11 H2O (water) + mixture water and acid

The process is actually the dehydration of sugar. It can be demonstrated by simply adding table sugar to a beaker and adding sulfuric acid to it. The process is followed by the release of heat, oxide fumes, and a black tube structure as mentioned. The white sugar on addition of this sulfuric acid get dehydrated and pushes itself outwards by forming a black carbon containing tube like structure.

Since the process is irreversible is nature therefore is a chemical change.

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