how many grams of zinc are needed to react fully with 8.0 moles of silver nitrate?

d. 0.599 moles of sulfur

To find the number of moles, you need to use the molar mass of sulfur, which is 32.06 g/mol.

First, divide the given mass by the molar mass to find the number of moles:

53.7 g / 32.06 g/mol = 1.674 moles

Then round to the appropriate number of significant figures, which is three in this case:

1.674 ≈ 0.599 moles

What is molar mass ?

The molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. Molar mass is a bulk not molecular, property of a substance.

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To determine the mass of pure aluminum that can be extracted from 655g of aluminum oxide, we need to use the balanced chemical equation for the reaction of aluminum oxide and carbon:

Al2O3 + 3C → 2Al + 3CO

This equation tells us that for every 1 mole of aluminum oxide that reacts with 3 moles of carbon, we will produce 2 moles of aluminum. We can use this information to convert the mass of aluminum oxide to moles of aluminum:

1. Calculate the molar mass of aluminum oxide (Al2O3):

2(Al) + 3(O) = 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol

2. Calculate the number of moles of aluminum oxide in 655g:

655 g / 101.96 g/mol = 6.42 mol

3. Use the mole ratio from the balanced equation to calculate the number of moles of aluminum that can be produced:

2 moles Al / 1 mole Al2O3 = 2 * 6.42 mol = 12.84 mol Al

4. Calculate the mass of aluminum produced:

12.84 mol Al x 26.98 g/mol = 346.2 g Al

Therefore, the mass of pure aluminum that can be extracted from 655g of aluminum oxide is 346.2 grams.

Since the given mass has only 3 significant figures, round this answer to 108 g

The pH of a 25.0 mL sample of a 0.3000 M solution of aqueous trimethylamine after 10.0, 20.0, and 30.0 mL of a 0.3750 M solution of HCl are 9.98, 11.46, and 13.24, respectively.

The pH of HCl can be calculated as follows:

After adding 10.0 mL of HCl, the amount of trimethylamine left is 0.00750 moles, and the amount of HCl added is 0.00375 moles. Since HCl is a strong acid, it will completely dissociate in water, so the amount of H⁺ ions added is also 0.00375 moles. The amount of trimethylammonium ions formed will also be 0.00375 moles. Using an ICE table and the Kb value of trimethylamine, the concentration of hydroxide ions is found to be 1.04×10⁻⁴ M, which gives a pH of 9.98.

After adding 20.0 mL of HCl, the amount of trimethylamine left is 0.00450 moles, and the amount of HCl added is 0.00750 moles. The amount of H⁺ ions added is now 0.01125 moles, and the amount of trimethylammonium ions formed is also 0.01125 moles. Using an ICE table, the concentration of hydroxide ions is found to be 3.47×10⁻⁶ M, which gives a pH of 11.46.

After adding 30.0 mL of HCl, the amount of trimethylamine left is 0.00150 moles, and the amount of HCl added is 0.01125 moles. The amount of H⁺ ions added is now 0.01406 moles, and the amount of trimethylammonium ions formed is also 0.01406 moles. Using an ICE table, the concentration of hydroxide ions is found to be 5.80×10⁻⁸ M, which gives a pH of 13.24.

Trimethylamine is a weak base, so it reacts with HCl to form the conjugate acid, trimethylammonium chloride, and water. As the amount of acid added increases, the concentration of hydroxide ions decreases, and the pH of the solution decreases, becoming more acidic.

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When the change in system entropy contributes to an increase in reaction spontaneity, the sign of the Gibbs free energy change (ΔG) is negative.

The Gibbs free energy change is given by the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy. For a reaction to be spontaneous, the Gibbs free energy change must be negative (ΔG < 0).

If the change in system entropy (ΔS) is positive, it means that the system has become more disordered or more random, which usually indicates an increase in the number of available states for the system. This increase in entropy tends to contribute to a decrease in the Gibbs free energy, making the reaction more spontaneous.

Therefore, when the change in system entropy contributes to an increase in reaction spontaneity, ΔS is positive, and the sign of ΔG is negative.

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When transitions between the energy states of a particular atom are E₁-> E₂-> E₃-> E₄ ->E₅, the transition that rise to a photon of longer wavelength is E₄ to E₃.

When an electron jumps from a large energy level to a smaller one, a photon is emitted, which has an energy equal to the specific energy difference between those respective two levels. The greater the energy difference between the two specific levels, the greater the photon's energy will be. The photon's energy is equal to the difference in the specific energy levels that are involved in the transition. Eph = ΔE, where Eph --> the energy of the photon and

The energy of the photon can be calculated as [tex]E_{ph}= \frac{h}{cλ}[/tex]

Now, [tex]\frac{h}{cλ} = ΔE[/tex]

=> [tex]\frac{6.626×10⁻³⁴}{3×10⁸∆E}= \lambda[/tex]

=> [tex]\frac{1}{∆E}= \lambda[/tex]

From the above expression, we can say that the less the energy difference between the levels of transition, the longer will be the wavelength of the photon emitted. So, energy difference is large in E₄ to E₃.

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Complete question:

A. E5 to E1.

B. E4 to E3

C. E3 to E1.

D. E4 to E1

The precipitate formed is (C) AgCl

The precipitate after the addition of 6 M HCl, then [tex]H_{2}S[/tex], and 0.2 M HCl to a sample containing [tex]Ba_{3}(PO_{4})_{2}[/tex], CdS, AgCl, [tex]NH_{4}Cl[/tex], and ZnS can be identified as follows:
1. After the addition of 6 M HCl:
- AgCl will precipitate due to its low solubility in HCl.
- [tex]NH_{4}Cl[/tex] will dissolve since it is a soluble salt.
2. After the addition of [tex]H_{2}S[/tex]:
- CdS and ZnS will precipitate since they form insoluble sulfides in the presence of [tex]H_{2}S[/tex].
3. After the addition of 0.2 M HCl:
- No new precipitates will form since the conditions have not changed significantly from the initial 6 M HCl addition.

So, the precipitates formed are AgCl, CdS, and ZnS. However, only AgCl is in the answer choices, making the correct answer C) AgCl.

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Hooke's law dictates that stretching frequencies are dependent on (a) Bond strength and molar masses of the atoms is the correct answer Option A.

Hooke's law states that the stretching frequencies of chemical bonds depend on the bond strength and the masses of the atoms involved. This law describes the relationship between the vibrational frequency of a bond and the force constant of that bond. So, if the bond is stronger, the stretching frequency will be higher, and if the atoms are heavier, the stretching frequency will be lower. The other options, such as the number of lone pairs or dipole moment, effective nuclear charge and polarizability, and magnetic spin and hybridization of the atoms, are not directly related to Hooke's law and the stretching frequencies of bonds.

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The process used to create a certain product is known as a synthetic pathway. Chemists occasionally need to be able to look backwards to examine the specific ingredients from which a desired product might be made.

The field of chemical science known as synthetic chemistry focuses on creating new chemical compounds and improving the processes used to create already existing ones. The participation of synthetic chemists in environmental chemistry is a crucial component of green chemistry.

In a chemical route, enzymes can be engaged at every stage. Because a certain enzyme is present at each stage, the molecule changes into a different form. A similar reaction pathway can either produce a new molecule (biosynthesis) or break down an existing one.

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Carbon dioxide gas, CO₂, effuses through a barrier at a rate of 0.40 ml/minute. if an unknown gas effuses through the same barrier at a rate of 0.31 ml/minute, the molar mass of the gas is 26.44g

Rate of a gas depends upon the molar mass of the given gas molecules in the inverse proportion of its square root.

It is related as mentioned below:

Given,

Rate of carbon dioxide gas= 0.40 ml/min

Rate of unknown gas =0.31 ml/min

Molar mass of carbon dioxide = 44 g/mol

On substituting the given value in the formula:

Rate of carbon dioxide / Rate of unknown gas

=[tex]\sqrt{molar mass of carbon dioxide/ molar mass of unknown gas}[/tex]

⇒0.40/0.31= [tex]\sqrt{molar mass of carbon dioxide/molar mass of unknown gas}[/tex]

Molar mass of unknown gas = 26.44g

Hence, the Molar mass of unknown gas is 26.44 g.

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Polymers are compounds that have the same molecular formula, but different, structure.

The various polymer structures can have a profound impact on the characteristics and behaviors of the polymer. Monomers, which are repeating units that make up polymers, can be joined by covalent bonds to form lengthy chains.

The properties of the polymers are significantly affected by the number of monomers, the branches in polymers, the linking fashion. For instance, the polymer polyethylene can be made from ethylene and its derivative, propylene.

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Complete question - Polymers are compounds that have the same molecular formula, but different,

The mass of sucrose when CO₂ is produced is 335.764 g which is shown below.

At STP, the value of temperature and pressure are 1.09 atm and 273.15 K.

To calculate the grams of CO₂, we will first calculate the number of moles of Co2 using the ideal gas equation which is expressed as follows-

PV = nRT

1.09 atm x 157 L= n  x 0.0821 L atm/mol/K x 273.15 K

n = 1.09 atm x 157 L / (0.0821 L atm/mol/K x 273.15 K)

  = 171.13 / 22.425

  = 7.631 moles

Now, using the below formula, the amount of CO₂ is grams can be calculated as follows-

It is know, the molar mass of CO₂ = 44 g/mol

n = m / M

7.631 moles = m / 44 g/mol

m = 335.764 g

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The method of food production that is not destructive to pantothenic acid is gentle cooking methods such as steaming, boiling, and microwaving.

These methods help preserve the nutrient content in the food, including pantothenic acid. On the other hand, high-temperature cooking methods such as frying, roasting, and baking can lead to the destruction of pantothenic acid.
Pantothenic acid, also known as vitamin B5, is sensitive to heat and can be destroyed during cooking or processing methods. By consuming raw or minimally processed foods, you can help preserve the pantothenic acid content in your food.

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Unknown substance gives a yellow color after performing the iodoform test, it confirms the presence of a methyl ketone group in the molecule. Therefore, the specific functional group present in the unknown is a methyl ketone (CH3-CO-).

If the unknown substance is found to be soluble in water, this indicates that it contains polar or ionic functional groups that can interact with the water molecules. Furthermore, a positive 2,4-DNP test indicates the presence of a carbonyl group in the unknown. This test detects the formation of a yellow-orange precipitate when the 2,4-DNP reagent reacts with the carbonyl group, which confirms the presence of this functional group.

The iodoform test is another qualitative test that can help identify the presence of a specific functional group in the unknown substance. This test is used to detect the presence of a methyl ketone (CH3-CO-) or an alcohol (-OH) group in the molecule. When a compound containing a methyl ketone or alcohol group reacts with iodine and sodium hydroxide, it produces iodoform (CHI3) as a yellow precipitate. Considering these facts, the specific functional group present in the unknown substance must be a methyl ketone.

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The pOH of the solution is 6.38 (Option B).

1. First, calculate the pH of the solution. The pH is the negative logarithm of the H₃O⁺ concentration. Use the formula:
  pH = -log[H₃O⁺]

2. Plug in the given H₃O⁺ concentration:
  pH = -log(2.4 × 10^-8)

3. Calculate the pH:
  pH ≈ 7.62

4. Next, calculate the pOH using the relationship between pH and pOH at 25°C:
  pH + pOH = 14

5. Solve for pOH:
  pOH = 14 - pH

6. Plug in the calculated pH:
  pOH = 14 - 7.62

7. Calculate the pOH:
  pOH ≈ 6.38

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The modern periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic numbers. The elements are arranged in the increasing order of their atomic numbers.

Here the element Lithium is a alkali metal which is located in the group 1 of the periodic table and placed in the second period. Its atomic number is 3. Alkali metals are highly electropositive and are good reducing agents.

Lithium reacts with the elements of groups 16 and 17 and form compounds, they also burn in oxygen and react with water.

The reaction of lithium with water is:

2Li + 2H₂O → 2LiOH + H₂

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The pH of the buffer solution is approximately 3.16, which corresponds to answer choice C.

A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it. This is due to the presence of a weak acid and its conjugate base (or a weak base and its conjugate acid) that can react with the added acid or base, thus maintaining the pH of the solution. In this case, the buffer contains a weak acid, HF, and its conjugate base, NaF.

To calculate the pH of the buffer, we need to use the Henderson-Hasselbalch equation, which is: pH = pKa + log([conjugate base]/[weak acid]).

First, we need to find the pKa of HF using the given Ka value:

[tex]Ka = [H+][F-]/[HF]

pKa = -log(Ka)

pKa = -log(3.5 × 10^-4)

pKa = 3.46

Now we can plug in the values for the buffer:

[conjugate base] = 0.040 M

NaF [weak acid] = 0.080 M HF

pH = 3.46 + log(0.040/0.080)

pH = 3.46 - 0.301

pH = 3.16[/tex]

Therefore, the pH of the buffer is C) 3.16.

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Alkyl halides capable of forming a carbocation are primary (1°), secondary (2°), and tertiary (3°) alkyl halides. Tertiary alkyl halides form the most stable carbocations, followed by secondary and primary alkyl halides.

Acetoacetic ester synthesis creates methyl ketone from acetoacetic ester and an alkyl halide by using both -carbon alkylation and -dicarboxylic acid decarboxylation. Because the reaction occurs via 2, a methyl halide or primary halide must be utilised in this kind of reaction. In essence, the acetoacetic ester's carbonyl group—which has two carbons on its side—connects with the alkyl group of the alkyl halide to create the methyl ketone.

The creation of a carbocation intermediate, which only involves the alkyl halide, is the reaction's rate-determining step in this kind of reaction. This is due to the fact that the dissociation of the alkyl halide into a carbocation and a halide ion occurs as the first step in a Sn1 reaction. The frequency of this dissociation depends on the concentration of the alkyl halide, which also affects the reaction's overall pace.

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The balanced equation for the oxidation of copper metal to Cu(II) and the reduction of silver ion to silver metal in a voltaic cell can be written as follows:

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

In this reaction, copper metal is oxidized to Cu2+ ions, while silver ions are reduced to silver metal. The overall reaction is spontaneous, and the electrons flow from the copper electrode (the anode) to the silver electrode (the cathode), generating an electrical current.

A voltaic cell is an electrochemical cell that converts chemical energy into electrical energy. It consists of two half-cells, each containing a metal electrode and a solution of ions.

The half-cell where oxidation occurs is called the anode, while the half-cell where reduction occurs is called the cathode. The two half-cells are connected by a salt bridge, which allows ions to move between the two solutions while keeping them electrically neutral.

In the case of the Cu-Ag voltaic cell, the anode is the copper electrode, and the cathode is the silver electrode.

Copper atoms at the anode lose electrons and are oxidized to form copper ions, while silver ions at the cathode gain electrons and are reduced to form silver atoms. The flow of electrons from the anode to the cathode generates a current that can be harnessed to do work.

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The pH of the solution that will results from the adding 150 ml of the 0.200 M HCl to the 150 ml of the 0.350 M NH₃ is 9.3.

The chemical equation is as :

HCl + NH₃  --->  NH₄⁺   +  Cl⁻

The moles of the HCl = molarity × volume

The moles of the HCl = 0.15 × 0.200

The moles of the HCl = 0.03 mol

The moles of the NH₃  = 0.15 × 0.350

The moles of the NH₃ = 0.0525 mol

Remaining moles of NH₃ = 0.0525 - 0.03

Remaining moles of NH₃ = 0.0225 mol

Total volume = 0.3 L

pOH = pH + log( acid/ base )

pOH = - log ( 1.8 × 10⁻⁵)

pOH = 4.7

pH = 14 - pOH

pH = 14 - 4.7

pH = 9.3

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Answer:

whats the question even ??/?

Explanation:

Intraspecific competition has a beneficial affect due to factors such as predators, disease and overpopulation. Thus option A is correct since all the above are beneficial factors.

Competition is described as an interaction between two or more individuals from the same or two or more populations wherein one negatively impacts the other in access to a limited resource (or resources) (food, water, nesting places, shelter, mates, etc.). Intraspecific competition occurs when individuals from the same species (cospecifics) compete. The impact of competition on each individual within the species is determined by the sort of competition that occurs.  When a species competes for a finite resource, all individuals eat equal quantities until the supply is gone, all members of that population may die of starvation.  On the other hand, when one individual competes and wins over a resource, and that resource is exploited, it continues to exist.

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If 7.78 g of cuso4 (molar mass 159.61 g/mol) is dissolved in water to make a 0.250 m solution of cuso4, the volume of the solution is 196 ml.

The molarity of a solution can be calculated using the following formula:

Molarity = no of moles of solute/volume of the solution in L

According to this question, a solution consists of 7.78 g of CuSO₄ in 0.250 M solution

no.of moles of CuSO₄ = mass of CuSO₄ /molar mass of CuSO₄

⇒ Molarity × volume of solution = mass of CuSO₄ /molar mass of CuSO₄

⇒ 0.250 M × volume of solution = 7.78g /159.61 gmol⁻¹

⇒volume of solution= 0.049 mol/0.250 molL⁻¹

⇒volume of solution= 0.196 L

⇒volume of solution= 196 ml

Therefore, volume of the  CuSO₄ solution is 196 ml

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The anticoagulant additive in a lavender-stoppered evacuated blood collection tube are sodium, lithium, or ammonium heparin. Option D is correct.

A lavender-stoppered evacuated blood collection tube typically contains an anticoagulant additive, which prevents blood from clotting and preserves the blood sample for further testing. The anticoagulant used in a lavender-stoppered tube is typically sodium, lithium, or ammonium heparin.

Sodium, lithium, or ammonium heparin works by inhibiting the activity of clotting factors in the blood, preventing the formation of clots and ensuring that the blood remains in a liquid state for testing purposes.

However, sodium fluoride and potassium oxalate are additives used in gray-stoppered tubes for glucose and alcohol testing, ethylene diamine tetra acetic acid (EDTA) is an additive used in purple-stoppered tubes for complete blood count (CBC) testing.

Hence, D. is the correct option.

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--The given question is incomplete, the complete question is

"What is the anticoagulant additive in a lavender-stoppered evacuated blood collection tube? multiple choice A) sodium fluoride and potassium oxalate B) ethylene diamine tetra acetic acid C) there are no additives. D) sodium, lithium, or ammonium heparin."--

To find the standard cell potential (E°cell) for the given cell reaction, we need to use the reduction half-reaction and the oxidation half-reaction and their respective standard reduction potentials (E*).

In this case, the reduction half-reaction is:
Sn4+(aq) + 2e- --> Sn2+(aq); E* = 0.15 V
And the oxidation half-reaction is:
2Al(s) --> 2Al3+(aq) + 6e- ; E* = -1.66 V
To balance the electrons in both half-reactions, we need to multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2.
3(Sn4+(aq) + 2e- --> Sn2+(aq)) --> 3Sn2+(aq) + 6e-
2(2Al(s) --> 2Al3+(aq) + 6e-)
Now, we can add these two half-reactions together to obtain the overall cell reaction:
2Al(s) + 3Sn4+(aq) --> 3Sn2+(aq) + 2Al3+(aq)
The standard cell potential (E°cell) can be calculated using the formula:
E°cell = E°reduction + E°oxidation
where E°reduction = E* of the reduction half-reaction and E°oxidation = -E* of the oxidation half-reaction.
Substituting the values:
E°cell = (0.15 V) + (-(-1.66 V))
E°cell = 1.81 V
Therefore, the answer is E) 1.81 V.

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Hi, to determine the E°cell for the cell reaction 2Al(s)+Sn4+(aq)--> 3Sn2+(aq)+2Al3+(aq), you need to follow these steps:

1. Write down the half-reactions:
  Oxidation: Al3+(aq) + 3e- <--> Al(s); E° = -1.66 V
  Reduction: Sn4+(aq) + 2e- <--> Sn2+(aq); E° = 0.15 V

2. Balance the number of electrons transferred in both half-reactions:
  Oxidation: 2(Al3+(aq) + 3e- <--> Al(s)); E° = -1.66 V
  Reduction: 3(Sn4+(aq) + 2e- <--> Sn2+(aq)); E° = 0.15 V

3. Calculate the E°cell by adding the E° values of the oxidation and reduction half-reactions:
  E°cell = E°(reduction) - E°(oxidation)
  E°cell = 0.15 V - (-1.66 V)
  E°cell = 1.81 V

Your answer: E) 1.81 V

If Peggy was mixing a fertilizer solution of Calcium nitrate, She uses 54. 0 g of [tex]Ca(NO_3)_2[/tex] and 300 ml of water, the molarity of the solution is

The molarity of the solution is defined as the number of moles of solute per liter in the solution. Molarity is a concentration term. The S.I. unit of mol/L of molarity.

Molarity = [tex]\frac{n}{V}*1000[/tex]

where n is the number of moles

V is the volume of solution (in milliliters)

Number of moles = [tex]\frac{m}{M}[/tex]

where m is the given mass

M is the molar mass

Molar mass of [tex]Ca(NO_3)_2[/tex]  = 164 g

Number of moles = [tex]\frac{54}{164}[/tex]

= 0.33

Molarity = [tex]\frac{0.33}{300}*1000[/tex]

= 1.1 mol/L

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The amount of energy in calories needed to convert 12 g of ice from -10*C to water at 50.0*C is 0.719311663 kilocalories.

To begin with, we must ascertain how much heat the water has absorbed. The energy conservation principle must then be used to calculate the heat that the pan releases.

Using this formula:

Q = m * c * ΔT

As follows:

m = 12 g = 0.012kg (water mass).

The specific heat capacity of water is 4.18 kJ/kg°C.

ΔT = (-10°C - 50°C) = -60°C

Q = 0.012 kg x 4.18 kJ/kg°C x 60°C

Q =3.0096 kJ or 0.719311663 KC

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Balancing chemical equations is essential to ensure the conservation of mass and charge in chemical reactions involving chemical decomposition and organic compounds.

To write and balance chemical equations, you must first identify the type of reaction.
(A) Combination/Synthesis: Two or more reactants combine to form a single product.
Example: A + B → AB
(B) Decomposition: A single reactant breaks down into two or more products.
Example: AB → A + B
(C) Replacement/Single Replacement: An element in a compound is replaced by another element.
Example: A + BC → AC + B
(D) Double Displacement: Two compounds exchange ions to form two new compounds.
Example: AB + CD → AD + CB
(E) Combustion of an Organic Fuel: An organic compound reacts with oxygen to produce carbon dioxide and water.
Example: [tex]C_{X} H_{Y}[/tex] + [tex]O_{2}[/tex] → [tex]CO_{2}[/tex] + [tex]H_{2} O[/tex]
In the context of your question, the term "decomposition" refers to reaction type (B), which involves the breakdown of a single compound into two or more products. The term "chemical" refers to the substances involved in these reactions, and "organic" is relevant to reaction type (E), where an organic compound undergoes combustion.

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The molar solubility of [tex]Ni(OH)_2[/tex] at a pH of 8.0 is 3. 8 mol/L.  

Molar solubility is the amount of a substance that is dissolved in a given amount of solvent (in this case, water) at a given temperature, and is typically measured in moles per liter (moles/L) or grams per liter (g/L). To calculate the molar solubility of  [tex]Ni(OH)_2[/tex] at a pH of 8.0, we need to know the concentration of the  [tex]Ni(OH)_2[/tex] in the solution, which is given in moles per liter (M).

The molar solubility of a substance can be calculated using the following formula:

Molar solubility (M) = concentration x solubility product constant

The solubility product constant for  [tex]Ni(OH)_2[/tex] is typically given in units of moles per liter (M/L) and is typically reported in literature as log K.

To calculate the molar solubility of  [tex]Ni(OH)_2[/tex] at a pH of 8.0, we need to know the concentration of the [tex]Ni(OH)_2[/tex] in the solution, which is given in moles per liter (M). We also need to know the solubility product constant for [tex]Ni(OH)_2[/tex], which is typically given in units of moles per liter (M/L) and is typically reported in literature as log K.

The molar solubility of [tex]Ni(OH)_2[/tex], can then be calculated using the following formula:

Molar solubility (M) = concentration x solubility product constant

For example, if the concentration of [tex]Ni(OH)_2[/tex] in the solution is 1 mol per liter (M) and the solubility product constant is log K = 3. 8, the molar solubility of [tex]Ni(OH)_2[/tex] at a pH of 8.0 would be:

Molar solubility (M) = 1 mol/L x 3. 8 M/L = 3. 8 M/L

Therefore, the molar solubility of [tex]Ni(OH)_2[/tex] at a pH of 8.0 is 3. 8 mol/L.  

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Correct Question:

Calculate the molar solubility of Ni(OH)2 when buffered at pH =8. 0.

The pH of a 0.023 M [tex]HNO_{3}[/tex] solution is approximately 1.64 (Option C).

To determine the pH of a 0.023 M [tex]HNO_{3}[/tex] solution, we need to consider the following terms: concentration (Molarity), pH, and the ionization of [tex]HNO_{3}[/tex] . [tex]HNO_{3}[/tex]  is a strong acid, meaning it fully ionizes in water.

Step 1: Identify the concentration of [tex]HNO_{3}[/tex] , which is given as 0.023 M.
Step 2: Since [tex]HNO_{3}[/tex]  is a strong acid, it will completely dissociate into H+ and NO3- ions. The concentration of H+ ions will be equal to the concentration of [tex]HNO_{3}[/tex] , which is 0.023 M.
Step 3: Use the pH formula: pH = -log[H+]. In this case, pH = -log(0.023).
Step 4: Calculate the pH: pH ≈ 1.64.

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