Balance the chemical equation
Fe2O3 (s) + CO (g) 2 Fe(s) + CO2 (g)
Express your answer as a chemical equation. Identify all of the phases in your answer.

Answer:

21.88mL is the volume of base required for the titration.

Explanation:

For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.

If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.

In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:

23.82mL - 1.94mL

Answer:

Kinetic energy = 1/2mv²

where m is the mass

v = velocity

m = 5.31 × 10^-23 kg

v = 1.00 × 10^2 m/s

Kinetic energy = 1/2 × 5.31 × 10^-23 × ( 1.00 × 10^2)²

= 2.655 × 10^-19 Joules

Hope this helps

Answer: There are [tex]2.29\times 10^{23}[/tex] formula units

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{36.0g}{95g/mol}=0.38moles[/tex]

1 mole of [tex]MgCl_2[/tex] contains = [tex]6.023\times 10^{23}[/tex] formula units

Thus 0.38 moles of [tex]MgCl_2[/tex] contains = [tex]\frac{6.023\times 10^{23}}{1}\times 0.38=2.29\times 10^{23}[/tex] formula units

Thus there are [tex]2.29\times 10^{23}[/tex] formula units

Answer:

The total surface area of these N spherical droplets is 4.4929 m²

Explanation:

From the information given :

assuming that :

30 cm³ of gasoline is atomized into N spherical droplets &

each with a radius of 2.0 × 10−5 m

We are tasked to determine the total surface area of these N spherical droplets

We all known that:

[tex]1 \ cm^3 = 10 ^{-6} m^3[/tex]

Therefore

[tex]30 \ cm^3 = 30 * 10 ^{-6} m^3 = 3 *1 0^{-5} \ m^3[/tex]

For each droplet; there is a required volume which is = [tex]\dfrac{4}{3} \pi r ^3[/tex]  since it assumes a sphere shape .

Thus;

replacing radius(r) with 2.0 × 10−5 m; we have:

[tex]= \dfrac{4}{3} \pi * (2.0 *10^{-5} m) ^3[/tex]

= [tex]3.35 * 10^{-14} \ m^3[/tex]

However; there are [tex]3*10^{-5} \ m^3[/tex] gasoline atomized into N spherical droplets with each with radius 2.0 × 10−5 m

For N ; we have ;

[tex]=\dfrac{3*10^{-5} \ m^3}{3.35 * 10^{-14} \ m^3/ droplet}[/tex]

= [tex]8.95*10^8 \ droplet s[/tex]

So; each droplet have a surface area = [tex]4 \pi r^2[/tex]

= [tex]4 \pi (2.0*10^{-5}m) ^2[/tex]

= [tex]5.02*10^{-9} \ m^2/droplets[/tex]

The surface area per droplet is equivalent to [tex]5.02*10^{-9} \ m^2/droplets[/tex]

Thus;

The total surface area of these N spherical droplets will be :

= [tex]8.95*10^8 \ droplet s * 5.02*10^{-9} \ m^2/ droplets[/tex]

= 4.4929 m²

The total surface area of these N spherical droplets is 4.4929 m²

Answer:

Path A-B-D involves a catalyst and is slower than A-C-D

Explanation:

The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.

The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.

The catalyzed path has a higher activation energy than the uncatalyzed path.

Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.

On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.

Answer:

Wavelength, λ  = 6.7 x 10^-11 m

Explanation:

Frequency and wavelength are inversely proportional to each other.

In this problem;

f =  4.5 x 10^18 Hz

wavelength, λ = ?

Speed of light, c = 3 x 108 m/s.

These variables are related by the following equation;

c = λ * f

Making λ subject of focus, we have;

λ = c / f

λ = 3 x 10^8 / 4.5 x 10^18

λ  = 0.67 x 10^-10

λ  = 6.7 x 10^-11 m

Answer:

[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]

Explanation:

[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]

Answer:

See explanation

Explanation:

For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:

2^n = 2^2 = 4 isomers

And the isomers that we can have are:

1) (1R,2S)-4-methylcyclohexane-1,2-diol

2) (1S,2S)-4-methylcyclohexane-1,2-diol

3) (1S,2S)-4-methylcyclohexane-1,2-diol

4) (1S,2R)-4-methylcyclohexane-1,2-diol

See figure 1

For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.

See figure 2

I hope it helps!

Answer:

Alumunium

Explanation:

Alumunium = [Ne] 3s² 3p¹

Ne = [He]2s²2p⁶

He = 1s

Alumunium = 1s 2s²2p⁶3s² 3p¹

Answer:

D

Explanation:

Answer:

22.29%

Explanation:

Percent yield = experimental yield / theoretical yield * 100

= 12.89 / 57.82 * 100 = 22.29%

Answer:

[tex]=C_3H_4O_3[/tex]

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to  assume 100 g of material. Thus,

Mass C = 40.92 g.  Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g.  Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g.  Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula [tex]=C_3H_4O_3[/tex]

Answer:

The incorrect statement from the options is OXIDATION IS A "GAIN" OF ELECTRONS

Explanation:

Oxidation in a redox reaction is the loss of electrons. It is also the increase in the oxidation states of an atom or ion or atoms in a molecule. A redox reaction is a type of chemical reaction in which there is a transfer of electrons from an atom or ion to another resulting in a change in oxidation states of the substances involved. The reducing agent in the reaction is undergoes oxidation by losing electrons while the oxidating agent is reduced that is it gains electrons at the end of the reaction. The atom or ion from which electron is lost is said to be oxidized while the other atom or ion involved in the reaction is reduced.

Oxidation is also the combination with O atoms and it is always accompanied by reduction because oxidation forms a half of the whole redox reaction. A substance cannot be oxidized except it has reduced another substance by losing electrons to it.

Answer:

A, B and C are compounds

Explanation:

First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).

Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.

Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.

Hence;

CaCO3(s)--------> CaO(s) + CO2(g)

Answer:

Explanation:

Partial pressure of oil = mole fraction of oil x total pressure

mole fraction of oil = mole of oil / mole of water + mole of oil

= mole of oil = mass of oil / molecular weight of oil

= 20 / 100 = .2

mole of water = 80 / 18

= 4.444

mole fraction of oil =  .2 / .2 + 4.444

= .2 / 4.644

Partial pressure of oil = mole fraction of oil x total pressure

= (.2 / 4.644 ) x 760 mm

= 32.73 mm Hg .

Answer:

[tex]V_2 = 4.87 * 10^3[/tex]

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

[tex]PV = nRT[/tex]

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]

[tex]\frac{V}{T} = \frac{nR}{P}[/tex]

Represent [tex]\frac{nR}{P}[/tex] with k

[tex]\frac{V}{T} = k[/tex]

[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

At this point, we can solve for the required parameter using the following;

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]

Multiply both sides by 323

[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]

[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]

[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]

[tex]V_2 = 4.87 * 10^3[/tex]

Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]

Answer:

"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system

"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.

"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.

"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.

Explanation:

Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.

The shape of the enzyme's active site is a negative of what the shape of the substrate would be.

On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.

This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.

Answer:

Explanation:

a. change of colour:

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.

Eg: copper reactions with the elements

b. Evolution of gas:

A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.

Eg: ammonium hydroxide breaks down to water and ammonia gas.

c. Change of smell :

Production of an Odor Some chemical changes produce new smells.  ... The formation of gas bubbles is another indicator that a chemical change may have occured.

Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.

d. Change of state:

A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.

Eg: candle wax (solid) melts initially to produce molten wax (liquid)

plz mark as brainliest!!!!

Answer:

d) El agua pesada contiene mas elementos

Explanation:

La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.

Answer:

....................

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Explanation:ki

Answer:

Molecules = 7.5 × 10²³ molecules

Explanation:

Given:

Moles = 1.25 mol

Avogadro's No. = [tex]N_{A}[/tex] = 6.022 * 10²³

Required:

Molecules = ?

Formula:

Molecules = Moles × [tex]N_{A}[/tex]

Solution:

Molecules = 1.25 × 6.022 × 10²³

Molecules = 7.5 × 10²³ molecules

Answer:

a lower temperature to liquefy

Explanation:

Answer: [H2SO4] = 0.5M;

              [KOH] = 1M

Explanation: Molarity is the solution concentration defined by:

molarity = [tex]\frac{mol}{L}[/tex] or M

To determine the concentration of the mixture, find how many mols of each compound there are in the mixture:

50 mL = 0.05L

H2SO4

1 mol/L * 0.05L = 0.05mol

KOH

2mol/L * 0.05L = 0.1 mol

The mixture has a total volume of:

V = 50 + 50 = 100 mL = 0.1 L

The concentration of the resullting solutes:

[H2SO4] = [tex]\frac{0.05}{0.1}[/tex] = 0.5 M

[KOH] = [tex]\frac{0.1}{0.1}[/tex] = 1 M

Concentration of H2SO4 is 0.5M while for KOH is 1M.

Combining quantum theory with wave motion of atomic particles is: Wave Mechanics

Answer:

33.3% of Sn in the sample

Explanation:

The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.

Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:

2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.

As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:

9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺

As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g

And the percentage of Sn in the sample is:

1.00g / 3.00g ₓ 100 =

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

Answer:

1.5 L  

Explanation:

If the gas is contained in a steel tank, the volume will remain constant when the temperature changes.

The volume will be 1.5 L.

Answer:

The third reaction

(2NaOH + NiCL2 ---> 2NaCl + Ni(OH)2)

Explanation:

By definition, a precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined.

(Source: lumenlearning)

From the 4 options, we can eliminate the first and second one immediately because there is no formation of an insoluble salt.

Then, the last one can also be eliminated because even though there is insoluble solid formed, but it is not a salt, and, the reactants are not solutions too. In fact, the last one is a displacement reaction. A more reactive metal displaces a less reactive metal to form an ion.

Since the third reaction matches the definition of precipitation reaction, this is the answer.

Answer:

C. Potential energy

Explanation:

Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.

Answer:

Slow combustion

Spontaneous combustion

Explosive combustion

Explanation:

-Slow combustion reactions: Occurs at low temperatures. Cellular respiration is an example.

-Spontaneous combustion reactions: Occurs suddenly without an outside heat source. The heat source is the result of oxidation.

-Explosive combustion reactions: Involves an oxidizing agent.

hopefully this helped :3

Answer:

Three types are: Rapid Combustion, Complete Combustion, and Spontaneous Combustion.

Explanation:

Note: there are more types! This is just three random ones I picked to list. Hope this helps! :)

Answer:

E) Al³⁺

Explanation:

A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.

In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;

A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.

K + 2e⁻  ----> K⁻²

B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.

Fe⁺ + 2e⁻  ----> Fe

C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.

O²⁻ + 2e⁻  ---->  O⁴⁻

D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.

Ne + 2e⁻   ---->   Ne²⁻

E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.

Al³⁺ + 2e⁻   ---->   Al⁺

The number of moles of carbon atoms in 0.500 mol of ethane (C₂H₆) is equal to one mole.

A mole can be defined as a scientific unit that is utilized to calculate the quantities such as atoms, molecules, ions, or other particular particles. The mass of one mole of a given chemical element is atomic mass and that of 1 mole of a chemical compound is molar mass.

The number of entities found in one mole is equal to 6.023 × 10 ²³ which is known as Avogadro’s constant.

Given, the number of moles of C₂H₆ = 0.500 mole

One molecule of ethane has carbons = 2

One mole of ethane has moles of carbons = 2 moles

0.500 mol of ethane has moles of carbon atoms = 0.500×2 = 1 mol

Therefore, one mole of carbon atoms is present in 0.500 mol of ethane C₂H₆.

Learn more about the mole, here:

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