Write a chemical equation for KOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Write a chemical equation for NH4+(aq) showing how it is an acid or a base according to the Arrhenius definition.
Express your answer as a chemical equation. Identify all of the phases in your answer.

Explanation:

They are called or known as cations

Answer:

1. 17.5 g of CO₂

2. The limiting reactant is carbon (graphite), and its formula is C(graphite)

3. 3.7 g of O₂

Explanation:

First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:

Thus, we write the chemical equation:

C(graphite) + O₂(g) → CO₂(g)

The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).

Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:

Mw(C) = 12 g/mol

moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol

Mw(O₂) = 16 g/mol x 2 = 32 g/mol

moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol

Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):

stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂

actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂

We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).

The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).

Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:

moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂

Now, we convert the moles of CO₂ to mass by using the Mw:

Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g

Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.

Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:

remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂

Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :

mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g

Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.

In a REDOX titration, one specie is oxidized while the other is reduced. The concentration of methanol is 0.012  mol L-1. Methanol is the oxidizing agent while permanganate is the reducing agent.

The average titre value is; [tex]\frac{7.95 + 7.10 + 7.00 + 7.10}{4}[/tex] = 7.29 mL

Equation of the reaction is:

[tex]4MnO4- + 12H+ + 5CH3OH ----> 4Mn2+ + 11H2O + 5HCOOH[/tex]

Concentration of oxidizing agent = CA = ?

Concentration of reducing agent = CB = 0.0021 mol L-1

Volume of oxidizing agent =  VA= 7.29 mL

Volume of reducing agent = VB = 50.00 mL

Number of moles of oxidizing agent NA = 4

Number of moles of reducing agent NB = 5

Note that NA and NB are obtained from the balanced reaction equation

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA = 0.0021 mol L-1 * 50.00 mL * 4/7.29 mL * 5

CA= 0.012  mol L-1

For a comprehensive definition of redox titration see

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Answer:

electric charge

Explanation:

Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.

When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.

A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.

A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.

Answer:

1.1 × 10²¹ NH₃ molecules

Explanation:

From the given information:

We were being told that the number of the hydrogen (H) atoms present in the sample of NH3 = 3.3 × 10²¹ hydrogen.

However, it signifies that each molecule of ammonia harbors 3hydrogen (H) atoms.

Hence,  the number of  molecules of NH₃ present;

[tex]\mathsf{=\dfrac{3.3\times 10^{21}}{3} \ molecules \ of \ {NH_3}}[/tex]

= 1.1 × 10²¹ NH₃ molecules

Answer:

"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.

Explanation:

Thus the above is the correct answer.

Answer:

1)In organic chemistry, the bromine test is a qualitative test for the presence of unsaturation (carbon-to-carbon double or triple bonds), phenols and anilines. ... The more unsaturated an unknown is, the more bromine it reacts with, and the less coloured the solution will appear.

2)The equation for incomplete combustion of propane is: 2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat. If not enough oxygen is present for complete combustion, incomplete combustion occurs. The result of incomplete combustion is, once again, water vapour, carbon dioxide and heat. But it also produces carbon monoxide.

Explanation:

3)Propene, also known as propylene, is an unsaturated organic compound with the chemical formula {\displaystyle {\ce {CH3CH=CH2}}}. It has one double bond, and is the second simplest member of the alkene class of hydrocarbons. It is a colorless gas with a faint petroleum-like odor. 

Formula: C3H6

IUPAC ID: Propene

4)Margarines are chemically created during hydrogenation which, until January 1, 2006, relied upon trans fats to solidify their vegetable oils. Food companies have been exploring options for replacing trans fat in partially hydrogenated margarine.

Answer:

boiling point is the physial property of matter

The option that is showing the physical property of the matter is boiling point, the correct option is C.

A physical property is any measurable property whose value describes the state of a physical system.

Changes in a system's physical properties can be used to describe its transitions between momentary states. Physical properties are also known as observables. They do not have modal properties.

Physical properties include color, phase, odor, and boiling point. Since reactivity with oxygen is dependent on the chemical nature of the object, it is not a physical property.

A compound's physical property is one that can be observed and measured. The chemical composition of a compound is unaffected by a physical property.

Thus, the correct option is C.

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Explanation:

Sulfuric Acid (H2SO4) is a strong mineral acid that has is colorless when pure. This chemical is used as a chemical intermediate to manufacture other chemicals and cleaning metal surfaces. The formula for sulfuric acid is H2SO4. The molar mass of sulfuric acid is 98.07848 g mol.

Answer:

Its C  you can check

Answer:

B. only formed in laboratories

Explanation:

i know

Isotopes are only formed in Laboratories. hence, Option (B) is correct.

Each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element is known as Isotope.

Isotopes are two or more types of atoms that have the same atomic number and position in the periodic table, and that differ in nucleon numbers due to different numbers of neutrons in their nuclei.

Therefore, Isotopes are only formed in Laboratories. hence, Option (B) is correct.

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Answer:

It is difficult because of C

Explanation:

The correct answer is:Some substances keep the same molecular structure when they break down, and others do not.

Some substances change their form when they dissolve and some do not. Dissolution can sometimes be regarded as a sort of reaction between a chemical substance and water.

Usually, we can consider dissolution of a substance in water as a sort of chemical reaction for some substances. For instance, an ionic substance interacts with water to form ions. similarly, some salts become hydrolysed in water and give acidic/basic solutions as  result of that.

However, some substances do not interact with water upon dissolution. They rather remain as molecular entities because they are not composed of ions.

We can see that some substances keep the same molecular structure when they break down, and others do not keep the same molecular structure  when they dissolve hence it is difficult to classify dissolving as a physical or a chemical change.

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Answer:

Explanation:

Your mom

Answer:

[tex]\%diff=24.0\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the van der Waals' equation as shown below:

[tex]p=\frac{RT}{v-b}-\frac{a}{v^2}[/tex]

Thus, we secondly calculate the molar volume as:

[tex]v=\frac{0.8311L}{9.998mol} =0.083L/mol[/tex]

Then, we plug in the entire variables in the vdW equation to get such pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol-0.03219L/mol}-\frac{1.345L*atm/mol}{(0.08313L/mol)^2}\\\\p=615.2atm[/tex]

And the ideal gas pressure:

[tex]p=\frac{0.08206\frac{atm*L}{mol*K}*502.7K}{0.08313L/mol}\\\\p=496.2atm[/tex]

Finally, the percent difference:

[tex]\%diff=\frac{|496.2atm-615.2atm|}{496.2atm} *100\%\\\\\%diff=24.0\%[/tex]

Regards!

Answer:

White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively

Explanation:

I hope it will help you

Answer:

A. High electrical conductivity

Explanation:

solid silver isn't brittle, it has a high melting point, and its not a good insulator.

Answer:

a well well well all el well eto eto rup wp will rup ei well dll alsof well po team app app app well piz wmv epic enz

Explanation:

Following are the kinetic theory of gases postulates:

1) Space-volume to molecules ratio is negligible.

2)There is no force of attraction between the molecules at normal temperature and pressure. The force of attraction between the molecules build when the temperature decreases and the pressure increases.

3) There is large space between the molecules resulting in continuous motion.

4) The free movement of molecules results in collision which is perfectly elastic.

5) The molecules have kinetic energy due to random movement. But the average kinetic energy of these molecules differs with temperature.

6) Molecules exert pressure on the walls of the container.

Answer:

a mixture that doesn't really show the ingredients or things put into the material or food.

Answer:

number of moles of gas increases the volume also increases.

Explanation:

The given balanced chemical equation is:

[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]

The value of Kc at 445oC is 0.020.

[HI]=1.5M

[H2]=2.50M

[I2]=0.05M

The value of Qc(reaction quotient ) is calculated as shown below:

Qc has the same expression as the equilibrium constant.

[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]

Qc>Kc,

Hence, the backward reaction is favored and the formation of Hi is favored.

Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. Moles ethanol -Molar mass: 46.07g/mol-:

25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol

kg solvent:

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

Molality:

0.54265 moles ethanol / 0.075kg = 7.24m

b. Liters solution:

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L

Molarity:

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

Water, mercury chloride and nitrogen oxide.

Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.

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Answer:

c. 29 J

Explanation:

Step 1: Given data

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J

Answer:

a measure of concentration equal to the gram equivalent weight per liter of solution.

Explanation:

Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

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