What are the first and second net ionic equations for the addition of aqueous ammonia to aluminum nitrate.

B. The development of a shopping center along the riverbank is  the following human activities would result in the most impact on the flow of water in a river

By building dams to store water and by taking water for household, commercial, and agricultural uses, humans actively alter the dynamics of the water cycle. Water demand and supply are expected to be even more impacted by climate change.

Human development, including population increase, reliance on fossil fuels, urbanisation, intercontinental trade, and industrial and agricultural emissions, is typically the cause of environmental changes in rivers.

Acid rain allows nitrous oxide and sulphur dioxide from industries and power plants to infiltrate river systems. In certain places, sewage and wastewater are dumped into rivers. All creatures, from algae to vertebrates, are impacted by pollution when the pH of the water is lowered. As pH falls, biodiversity declines.

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Answer: a physical change is something we can see, but a chemical change is not always something that we can see.

Explanation:  Physical changes show a change in appearance.

There are chemical changes when two aqueous solutions are added together and the product is still two aqueous solutions, so we would not see any change and no precipitate is formed.  

So a physical change is something we can see, but a chemical change is not always something that we can see.  

Explanation:

We are given: volume = 2.55L

At STP: Temperature = 273.15 K

: Pressure = 1 atm

: R = 0.08205 L.atm/mol.K

Using the ideal gas law:

Answer:

n = 0.11 mol

Answer:

See image below

Explanation:

To convert the left structure to the right structure, one pair of electrons on the oxygen is pushed down onto the carbon-oxygen bond (resulting in a double bond). This results in too many electrons on the top carbon. The electrons from the adjacent double bond are subsequently pushed onto the second carbon.

a. The mass of MgSO₄ that can be produced in the given reaction is 1.985 grams.

b. The percent yield of MgSO₄ is 10.68%

The complete balanced chemical equation is,

Na₂SO₄(aq)+MgCl₂(aq)⟶MgSO₄(s)+2NaCl(aq)

We know,

Molarity(M) = number of moles(n)/per litres solution.

So,

Number of moles of Na₂SO₄ = Molarity × volume.

Moles of Na₂SO₄ = 0.190×0.0871

Moles of Na₂SO₄ = 0.016549 moles.

Moles of MgCl₂ = 0.311×0.144

Moles of MgCl₂ = 0.044473 moles.

As we can see, moles of Na₂SO₄ are present in less amount,

So Na₂SO₄ will be Limiting reagent and it will control the reaction.

So,

Moles of Na₂SO₄ = Moles of MgSO4

Moles of MgSO₄ = 0.016549 moles.

As we know,

Moles = Mass produced/molar mass

Molar mass of MgSO₄ = 120grams/mole

Putting the values,

0.016549 = mass of Mg produced/120

Mass of MgSO₄ produced = 1.985 grams.

b. Percentage yield formula,

Percentage yield = experimental mass/theoretical mass×100

Theoretical mass = 1.985 grams

Experimental mass = 0.212 grams.

Percentage yield = 0.212/1.985×100

Percentage yield of MgSO₄ = 10.68%

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The percentage of methane is 12.9% and percentage of water is 87.09% in methyl hydrate.

It is given that for every methane molecule, 6 moles of water is present. Then the chemical formula of methane hydrate would be CH₄.6H₂O.The molecular mass of CH₄.6H₂O would be,

Molecular mass of CH₄.6H₂O = 12 + 4 + 6(2 +16)

                                                 = 124g/mol

To find the mass percentage of Methane, the following formula is to be used:

Mass percentage of CH₄ = (Mass of methane / Molecular mass of methane hydrate)x100

On substituting the values of mass of CH₄ and mass of CH₄.6H₂O we get,

Mass percentage = (16/124)100

Mass percentage of CH₄ = 12.9%

Therefore the mass percentage of CH₄ in CH₄.6H₂O is 12.9%

Similarly, to find the mass percentage of 6H₂O in methyl hydrate, the following formula is used:

Mass percentage of water = (Mass of 6H₂O/Molecular mass of CH₄.6H₂O)x100

On substituting the values of mass of 6.H₂O and CH₄.6H₂O we get,

Mass percentage of 6H₂O = (108/124)x100

Mass percentage of 6H₂O = 87.09%

Therefore the mass percentage of water in methyl hydrate is 87.09%

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suppose you need to calibrate a calorimeter. you burn 0.744 g of a fuel with an enthalpy of combustion of -48.7 kJ/g. if the temperature of water in the calorimeter increases by 106 °C during the reaction . the heat capacity of the calorimeter is 3.417 kJ/ °C.

Heat capacity equation is given by :

q = c ΔT

given that :

change in temperature ΔT = 106 °C

now, q = 0.744 g × ΔH

            = 0.744 g × 48.7 kJ/g

        q = 36.23 kJ

putting the value in formula :

q = c ΔT

c = q /ΔT

c = 36.23 kJ / 10.6  °C

c = 3.417 kJ/  °C

Thus, suppose you need to calibrate a calorimeter. you burn 0.744 g of a fuel with an enthalpy of combustion of -48.7 kJ/g. if the temperature of water in the calorimeter increases by 106 °C during the reaction . the heat capacity of the calorimeter is 3.417 kJ/ °C.

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Answer: A mole of Aluminium Sulphate contains 12 moles of Oxygen

So

12

1

​

×0.24

Moles of Aluminium Sulphate contains 0.24 moles of Oxygen

So 2×10

−2

Moles is the answer

Explanation:

The final concentration of the solution is 0.13 M,

The term dilution has to do with a reduction in the concentration of a solution. We ought to note that the term concentration has to do with the amount of substance that we can find in a solution. There are several units that we could use in the measurement of the concentration of a solution and these include;

Now we are required to be able to obtain the concentration of the solution of the copper I ion that can be found in the flask that has only 100mL. Let us recall that the original solution had only 5 mL of it removed.

Given that;

C1V1 = C2V2

C1 = initial concentration

C2 = Final concentration

V1 = initial volume

V2 = final volume

C2 = C1V1/V2

C2 = 2.63M  * 5.0mL/100 mL

C2 = 0.13 M

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Answer: 1.4 mol A2B is produced.  

Explanation:  2A + B = A2B, so there are 2 moles of A to every one mole of B.  

So if 2.8 mol of A is present 1/2 that amount is made in A2B (and there will be extra B left over because A is the limiting reactant and you can't make more product than the least amount of reactant allows.  

So 1.4 mole of A2B will be produced.  

Answer and Explanation: A molecule consisting of central atom with three bond pairs and one lone pair of electrons will have trigonal pyramidal shape.

If you dissolved 54 grams of ammonia in 100 grams of water at 400°C then the solution is saturated

A saturated solution is a solution that has dissolved as much solute as it is capable of dissolving and in a saturated solution no more solute can be dissolved at a given temperature and we can make an saturated solution by keep on dissolving solute until no more solute can be dissolved

Here given data is dissolved 54 grams of ammonia in 100 grams of water at 400°C then the ammonia is totally dissolved in this 100 ml water

NH₃ + H₂O → NH₄OH

So, here ammonium hydroxide are formed which is saturated solution and when ammonia is dissolved in water the water molecule donate the proton to the NH₃ molecule and the solubility of ammonia decreases with the increases in of water solvent that's why when we dissolved  54 grams of ammonia in 100 grams of water at 400°C the solution is saturated

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Answer:

There would be 4 atoms in a tetrahedral molecule.

Explanation:

Answer: 4 atoms

Explanation:

The theoretical yield of aluminum iodide according to the reaction would be 276.5 grams.

We are looking to get the theoretical yield of aluminum iodide. This can be achieved by first getting the balanced equation of the reaction:

[tex]2Al + 3I_2 --- > 2AlI_3[/tex]

The mole ratio of Al that reacts and the [tex]AlI_3[/tex] that is produced is 1:1.

Recall that: mole = mass/molar mass.

18.3 grams of aluminum completely reacts:

Molar weight of aluminum = 26.98 g/mol

Mole of 18.3 g aluminum = 18.3/26.98

                                         = 0.6782 mole

Since the mole ratio of Al and  [tex]AlI_3[/tex] is 1:1, the equivalent mole of  [tex]AlI_3[/tex] will also be 0.6782 mole.

Recall that: mass = mole x molar mass

Molar mass of [tex]AlI_3[/tex] = 407.695 g/mol

Mass of 0.6782 [tex]AlI_3[/tex] = 0.6782 x 407.695

                                 = 276.5 grams

In other words, the theoretical yield of  [tex]AlI_3[/tex] according to the reaction is 276.5 grams.

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Acid-base reactions are reversible and which ones are irreversible then the 1st reaction is HA + H₂O ↔ A⁻ + H₃O⁺ is the reversible reaction and in the 2nd reaction MOH → M⁺ + OH⁻ is the reversible reaction and 3rd reaction is HA+H₂O → A⁻ + H₃O⁺ is the irreversible reaction

Acid base reaction is the chemical reaction that occur when acid and base are mixed together called acid base reaction and irreversible reaction means they can occur only in one direction the reactants can change to the products but the products cannot change back to the reactants and reversible reaction means chemical reaction occur in both direction the reactants can change to the products and the products can also change back to the reactants and in that given reactions

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The molarity of 245.0 g of H₂SO4 dissolved in 1.00 L of solution is 2.498M. the molarity of 5.00 g of NaOH in 750.0 mL of solution is 0.167M.

A chemical species' concentration in a solution, specifically the amount of a solute per unit volume of solution, is measured by its molar concentration.

The number of moles per liter when denoted by the unit symbol mol/L or mol/dm3 in SI units, is the mostly used unit used for denoting molarity in chemistry. The secret to determining molarity is to keep in mind that it is measured in moles per liter (M).

MV = grams / molar mass

(x)*(1.000 L) = (245.0 g) / (98.0768 g mol¯1)

x = 2.49804235 M

So, calculating it to 4 significant figures, x = 2.498 M

For 2nd question

Applying the same formula, MV = grams / molar mass

(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1

(x) (0.7500 L) = 0.1250097 mol

x = 0.1666796 M

x = 0.167 M (when calculated to three Significant figures)

For 3rd question,

using variation of the same formula,

0.200 M = x / 0.1000 L

x = 0.0200 mol

For the 4th question,

MV = moles of solute

no of moles x = (0.415 mol L¯1) (0.01600 L)

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The hill is 1.7 metres high (Approx.)

Knowing that;

Energy potential = 1000 J

Weight = 60 kg

mgh = potential energy

1,000 = [60][9.8][h]

h = 1.7 m

The whole amount of energy will be in the form of potential energy since the ball's kinetic energy will be equal to zero. As a result, when the ball reaches the peak of its climb, its potential energy will be 450 J. As a result, the ball will soar as high as 45.91 metres.

An enormous amount of potential or stored energy exists in a rock that is at rest on a hill. The boulder's potential energy transforms into kinetic energy, the energy of motion, when it falls.

The automobile drives faster as it goes down the hill, which results in an increase in kinetic energy and a decrease in potential energy.

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The complete ionic and net ionic forms of the following equation.

ionic equation :

Mn²⁺  +  2Cl²⁻  +  2NH₄⁺  +  CO₃²⁻   ----->   MnCO₃   +  2NH₄⁺ +  2Cl⁻

net ionic equation :

Mn²⁺(aq)   +   CO₃²⁻(aq)   ---->    MnCO₃(s)

according to question :

MnCl₂    +   (NH₄)₂CO₃   ----> MnCO₃     +    NH₄Cl

balanced molecular equation : with states

MnCl₂(aq)    +   (NH₄)₂CO₃(aq   ----> MnCO₃(s)     +    2NH₄Cl(aq)

the complete ionic equation as follows :

Mn²⁺  +  2Cl²⁻  +  2NH₄⁺  +  CO₃²⁻   ----->   MnCO₃   +  2NH₄⁺ +  2Cl⁻

Now cancel out the spectator ions on both side of equation , the reactant and the product, we get

Mn²⁺(aq)   +   CO₃²⁻(aq)   ---->    MnCO₃(s)

now, the complete net ionic equation is

Mn²⁺(aq)   +   CO₃²⁻(aq)   ---->    MnCO₃(s)

Thus, The complete ionic and net ionic forms of the following equation.

ionic equation :

Mn²⁺  +  2Cl²⁻  +  2NH₄⁺  +  CO₃²⁻   ----->   MnCO₃   +  2NH₄⁺ +  2Cl⁻

net ionic equation :

Mn²⁺(aq)   +   CO₃²⁻(aq)   ---->    MnCO₃(s)

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Answer:

OsBr2.

Explanation:

Osmium decobromide is the same as osmium bromide. This is the result of the reaction between osmium hydroxide (Os(OH)2) and hydrobromic acid (HBr) to form water and a salt (osmium bromide):

So, osmium bromide has the formula: OsBr2.

Answer:

Each 2 ml ampoule contains 4 mmol Mg2+ (equivalent to 1 g magnesium sulfate heptahydrate). Each 10 ml ampoule contains 20 mmol Mg2+ (equivalent to 5 g magnesium sulfate heptahydrate). The concentrations of magnesium ions (Mg2+) in millimoles are given as approximate values.

Energy is added in the conversion of solid to liquids and then the conversion of liquids to gases.

We know that matter occurs in three states; Solid, liquid and gas. Now, we know that solid matter has its particles held together in place such that they do not move about from place to place. The particles of a liquid are not held so closely together hence they do not translate to a large extent. The molecules of a gas are very much free and move about freely from place to place.

Now, we know a solid can be converted to a liquid when it is heated such that the molecules of the solid receive more kinetic energy and move faster and the rigid intermolecular forces that hold them together are broken down thus heat is added.

When we add more heat to the liquid, the intermolecular forces also break down and the we have the gaseous state in which there are no bonds between the molecules.

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help me i don't know

explain me hahaha

If 22.2 g of iodine reacts with excess hydrogen gas, the amount, in grams, of hydrogen iodide that would be formed is 44.75 grams.

From the balanced equation of the reaction, as shown in the illustration, the mole ratio of the iodine that reacts and the hydrogen iodide that forms is 1:2.

22.2 g of iodine reacted.

Recall that: mole = mass/molar mass

Molar mass of iodine = 126.9 g/mol

Mole of iodine that reacted = 22.2/126.9

                                          = 0.1749 moles

From the mole ratio, the equivalent mole of hydrogen iodide that will be produced is:

           0.1749 x 2 = 0.3499 moles

Recall that: mass = mole x molar mass

Molar mass of hydrogen iodide = 127.9 g/mol

Mass of 0.3499 mole hydrogen iodide = 0.3499 x 127.9

                                                             = 44.75 grams

Thus, the amount of hydrogen iodide that will be produced is 44.75 grams.

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The question requires us to calculate the molarity (or molar concentration) of a HNO3 solution, given the volume and concentration of a KOH solution used to neutralize a given volume of the acid.

The following information was provided by the question:

molarity of KOH solution = M(KOH) = 0.01 M = 0.01 mol/L

volume of KOH solution = V(KOH) = 49.90 mL

volume of HNO3 solution = V(HNO3) = 20.00 mL

To answer this question, we need to take a look at the neutralization reaction between HNO3 and KOH:

HNO3 + KOH -> KNO3 + H2O

From the reaction, we have that 1 mol of HNO3 is necessary to neutralize 1 mol of KOH. With this information, we can write that, to achieve complete neutralization:

number of moles of HNO3 = number of moles of KOH

Since the molarity of a solution is given by the following equation:

We can rearrange it to:

And we can also write the equality between the number of moles of HNO3 and KOH as:

And, since the concentration of KOH was provided, as well as the volumes of HNO3 and KOH necessary to achieve neutralization, we can apply these values to calculate the molarity of HNO3 using the equation above:

Therefore, the molarity of the nitric acid solution is equal to 0.02495 mol/L.

The smooth ER differs from the rough ER in that the smooth ER produces carbohydrates and lipids, whereas the rough ER produces proteins. (Option B)

The manufacture, folding, quality assurance, and shipment of some proteins take place in the rough ER, which is dotted with millions of membrane-bound ribosomes.

The production of steroid hormones, lipid (fat) synthesis, and metabolism are all heavily correlated with SER. It also serves as a detoxifier. The production of proteins is greatly aided by the rough endoplasmic reticulum.

The fundamental distinction between the rough endoplasmic reticulum and the smooth endoplasmic reticulum is that the RER contains ribosomes, which give it its rough appearance and enable protein synthesis.

Ribosomes cover the rough endoplasmic reticulum, which is important in the synthesis and creation of proteins.

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Answer:

Explanation:

yyyyyyyyyyyy

The volume of HCl is needed to completely neutralize is 81.25 ml.

Formula of neutralization, M₁V₁ = M₂V₂

M₁ = concentration of HCl

V₁ = volume of HCl

M₂ = concentration of NaOH

V₂ = volume of NaOH

Put the value in M₁V₁ = M₂V₂

0.100 × V₁ = 25 × 0.325

V₁ =  25 × 0.325/ 0.100

V₁ = 81.25 ml

Hence, the volume of HCl is 81.25 ml.

In chemistry, neutralization or neutralization is a chemical reaction wherein acid and a base react quantitatively with every other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present inside the answer.

whilst a strong acid reacts with a strong base the ensuing salt is neither acidic nor primary in nature i.e. it's far impartial for instance whilst HCl, a strong acid, reacts with NaOH, a robust base, the resulting salt is sodium chloride and water.

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nswer

Explanation

An exothermic reaction is a type of chemical reaction that gives off heat to its surroundings. In an exothermic reaction, energy is released because the total energy of the products is less than the total energy of the reactants.

Answer:An exothermic Chemical Reaction

Explanation:

The binding energy of a mole of nuclei with a mass defect of 0.00084kg/mol is 7.56 x 10¹³ J/mol (option B).

Binding energy is the energy needed to separate the constituent parts of an atom or nucleus; equivalent to the mass defect.

The binding energy of a nucleus can be calculated as follows:

Binding Energy = mass defect x c²

where;

According to this question, a nuclei with a mass defect of 0.00084kg/mol is given. The binding energy can be calculated as follows:

Binding energy = 0.00084 kg/mol × (2.9979 x 10⁸ )²

= 8.94e16 × 8.4 × 10-⁴

= 7.51 × 10¹³J

Therefore, 7.51 × 10¹³J/mol is the binding energy of the nuclei.

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The five systems that helps us grow from a child to adult are excretory, integumentary, digestive, respiratory and cardiovascular system.

The body systems are the combination two or more systems located in the body that helps in the maintenance of hemostasis of the body.

Examples of body systems include the following:

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