calculate the ph of a solution prepared from 0.198 molmol of nh4cn and enough water to make 1.00 ll of solutio

Matching the solutions with the appropriate letters, we have:

0.147 m CuSO4 - C

0.205 m Ni(NO₃)₂ - B

8.75 × 10⁻² m Cr(CH₃COO)₃ - D

0.380 m ethylene glycol - A

Based on the information provided, we need to match the given aqueous solutions with the appropriate letter from the column on the right. The options are:

A. Highest boiling point

B. Second highest boiling point

C. Third highest boiling point

D. Lowest boiling point

Let's analyze each solution and determine their boiling points:

0.147 m CuSO₄ (copper sulfate) - This is an ionic compound and will dissociate into Cu²⁺ and SO₄²⁻ ions in water. As an electrolyte, it will exhibit colligative properties, including an increase in boiling point. Therefore, this solution would have the third highest boiling point. So the match is C.

0.205 m Ni(NO₃)₂ (nickel nitrate) - Similar to the previous solution, this is also an ionic compound and will dissociate into Ni²⁺ and NO³⁻ ions in water. It will exhibit colligative properties, resulting in a higher boiling point. This solution would have the second highest boiling point. So the match is B.

8.75 × 10⁻² m Cr(CH₃COO)₃ (chromium(III) acetate) - This is also an ionic compound and will dissociate into Cr³⁺ and CH₃COO⁻ ions in water. Like the previous solutions, it will exhibit colligative properties, leading to an increase in boiling point. This solution would have the lowest boiling point. So the match is D.

0.380 m ethylene glycol - Ethylene glycol is a nonelectrolyte, and it does not dissociate into ions in water. Therefore, it does not exhibit colligative properties to the same extent as ionic compounds. However, it still has a significant effect on the boiling point due to its high boiling point itself. Ethylene glycol has the highest boiling point among the given options. So the match is A.

Matching the solutions with the appropriate letters, we have:

0.147 m CuSO₄ - C

0.205 m Ni(NO₃)₂ - B

8.75 × 10⁻² m Cr(CH₃COO)₃ - D

0.380 m ethylene glycol - A

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1. The bacterial cell shown is a prokaryotic cell.

2a. The similarities between prokaryotic and eukaryotic cells include:

2b. The differences between prokaryotic and eukaryotic cells include:

Eukaryotic cells are cells that possess a membrane-bound nucleus as well as membrane-bound cell organelles.

Eukaryotic cells are cells of higher organisms such as animals and plants.

Prokaryotic cells are cells that lack a membrane-bound nucleus as well as membrane-bound cell organelles.

Prokaryotic cells are cells of lower organisms such as bacteria.

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The new concentration of the solution is  0.85 mole/L.

We know that the term concentration has to do with the amount of the solute in a solution. In this case, we know that we had about  617. ml  of the solution and then we now happen to remove 253. ml  and we are left with about 364 mL.

Then;

Concentration = Number of moles/Volume of the solution

The new concentration of the solution can now be obtained as;

0.310 m/364 * 1/1000

= 0.85 mole/L

The solution would now have the concentration of about 0.85 mole/L.

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There are 23 atoms of Fe in a 55.845 gram. As a result, after dividing our mass of 239 grams by the iron formula mass of 55.84 grams per bole, formol equals after calculation.

Use the millimeters to mole formula to calculate the molecular weight n, of a substance with a given mass, m, (in grams) accurately. n = m / M, where M is the substance's molar mass also known as gram-molecular weight. a substance's molecular weight expressed as a mass in grams. Example: NaCl salt weighs 58.44 grams per gram-mole. American Meteorological Society, copyright 2022.

Number of Avogadro = 6.023 × 10²³. The products of every chemical reaction are measured using the Avogadro number. 1 mole of atoms, molecules, or particles is equal to 6.023 x 1023. Calculating the number of molecules is as follows: mole number = quantity of material / mass of one mole.

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There are 23 atoms of Fe in a 55.845 gram. As a result, after dividing our mass of 239 grams by the iron formula mass of 55.84 grams per bole, formol equals after calculation.

Use the millimeters to mole formula to calculate the molecular weight n, of a substance with a given mass, m, (in grams) accurately. n = m / M, where M is the substance's molar mass also known as gram-molecular weight. a substance's molecular weight expressed as a mass in grams. Example: NaCl salt weighs 58.44 grams per gram-mole. American Meteorological Society, copyright 2022.

Number of Avogadro = 6.023 × 10²³. The products of every chemical reaction are measured using the Avogadro number. 1 mole of atoms, molecules, or particles is equal to 6.023 x 1023. Calculating the number of molecules is as follows: mole number = quantity of material / mass of one mole.

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Fe's mass percent composition is equal to 55.85 g/mol times 329.27 g/mol, or 100%. Fe's mass percent composition is equal to 0.1696 times 100%. Fe has a mass percentage composition of 16.96%.

Mass percent is best expressed using the formula mass percentage  mass of chemical x measure the mass of combination) x 100. To express the amount as a percentage, multiply the value at the top by 100.

Titanium (35 percent), oxygen (30 percent), silicate (15 percent), and aluminium make up the majority of the Earth's mass (13 percent).

The mass percent is determined by dividing the amount of compound or solute by the amount of the component or solute. A percent is obtained by multiplying the result by 100. A compound's composition can be determined by applying the formula: mass percentage = (mass of element in 1 mole of compound /mass of 1 mole of compound ) 100.

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                                 "Complete question"

Lab Report Sheet: Part A: Determination of Mass Percent of Iron (Fe) Mass of evaporating dish and unknown sample _____ g Mass of evaporating dish _____ g Mass of original sample _____ g Mass of evaporating dish after removing iron fillings _____ g Mass of Fe _____ g Percent of Fe in sample _____ % Calculations: fill in the blanks.

The stabilization and destabilization of octahedral complexes refer to the changes in the energy levels of d-orbitals in transition metal complexes, which affects their properties and reactivity.

In octahedral complexes, the d-orbitals of the central metal ion split into two sets of energy levels due to the presence of ligands. This is known as crystal field splitting. The energy gap between these sets is determined by the strength of the ligand field, which is related to the nature of the ligands and the geometry of the complex.
Stabilization occurs when the ligand field is strong, causing a large energy gap between the two sets of orbitals (t2g and eg). This leads to lower energy and more stable complexes. Examples of strong ligands that cause stabilization include CN-, CO, and NO2-.
Destabilization, on the other hand, occurs when the ligand field is weak, causing a smaller energy gap between the sets of orbitals. This leads to higher energy and less stable complexes. Examples of weak ligands that cause destabilization include I-, Br-, and Cl-.
In summary, the stabilization and destabilization of octahedral complexes are determined by the ligand field strength and the resulting energy gap between the d-orbitals, affecting the properties and reactivity of the complexes.

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When 10.0 g of sodium react with 10 g of chlorine gas according to the equation, the mass of sodium chloride produced is 25.4 g.

The mass of sodium chloride produced in the given reaction is calculated as follows;

2Na + Cl2 = 2Nacl

In the reaction above;

2(23 gram of sodium) -------------------- 2(58.44 g of sodium chloride)

10 g of sodium -------------------------- ? mass of sodium chloride

46 g of Na ----------------------------116.88 g of NaCl

10 g of Na ---------------------------- ? NaCl

= (10 x 116.88)/46

= 25.4 g

Thus,  when 10.0 g of sodium react with 10 g of chlorine gas according to the equation, the mass of sodium chloride produced is 25.4 g.

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As the metal scale will shrink as a result of the heat, making the measurement inaccurate, it would be too high.

The chemical reaction between potassium chloride and silver nitrate, in which solid silver chloride is precipitated out, is one of the greatest examples of precipitation reactions. This is the precipitation reaction's byproduct, the insoluble salt. Below is given the chemical equation for this precipitation process.

AgNO3(aqueous) + KCl(aqueous) → AgCl(precipitate) + KNO3(aqueous)

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The mass of C₅H₅NHCl , the student should be dissolve in the solution to turn it into a buffer with ph = 5.64 is equals to the 31.45g.

We have a pyridine solution, which is a weak base solution. For a weak base,

the ionization constant, kb = 1.7 × 10⁻⁹

Buffer pH = 5.64

Volume of solution = 250 mL

Molarity of solution = 0.7 M

Using pH formula, pKb = - log ( 1.7 × 10⁻⁹)

= 8.77

For pH of equilibrium constant of water,

pKw = pKb + pKa

=> 14 = 8.77 + pKa

=> pKa = 5.23

Buffer pH formula = pKa + log( [[C₅H₅N] /[C₅H₅NHCl])

=> 5.64 = 5.23 + log( 0.7 M/[C₅H₅NHCl])

=> log(0.7 M/[C₅H₅NHCl]) = 5.64 - 5.23

=> log(0.7 M/[C₅H₅NHCl]) = 0.41

So, 0.7 M/[C₅H₅NHCl] = 10⁻⁰·⁴¹

=> [C₅H₅NHCl] = 10⁻⁰·⁴¹ × 0.7 M

Moles of [C₅H₅NHCl] = 250 mL× 10⁻⁰·⁴¹ × 0.7 mol [C₅H₅NHCl] / 1000 mL

= (0.7 ×10⁻⁰·⁴¹)/4

= 0.175 ×10⁻⁰·⁴¹ moles = 0.2723 moles.

Molar mass of C₅H₅NHCl = 115.5 g/mol

Mass of [C₅H₅NHCl] = 0.175 ×10⁻⁰·⁴¹ moles × 115.5 g [C₅H₅NHCl]/ 1 mol of [C₅H₅NHCl]

= 0.2723 × 115.5 g = 31.454 g

Hence, required mass is 31.45 g.

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Complete question:

a chemistry graduate student is given 250 ml of a 0.7 M pyridine solution C5H5N. pyridine is a weak base with Kb = 1.7 × 10-9. what mass of C5H5NHCl should the student dissolve in the solution to turn it into a buffer with ph= 5.64 ? you may assume that the volume of the solution doesn't change when the is dissolved in it. be sure your answer has a unit symbol, and round it to significant digits.

According to the stoichiometry of the chemical reaction,0.713 moles of water can be made when 134.9 grams of HNO₃ are consumed.

It is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample.

378.24 g nitric acid gives 36 g water, thus 134.9 g nitric acid gives 134.9×36/378.24=12.83 g which is 12.83/18= 0.713 moles.

Thus,0.713 moles of water can be made when 134.9 grams of HNO₃ are consumed.

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d.) Microwave photons cause the molecules to increase their rotational energy states, whereas infrared photons cause electrons in the molecules to increase their electronic energy states.

Explanation:

Microwave: transitions in the molecular rotational levels

Infrared: transitions in molecular vibrational levels

UV/Visible: transitions in electronic energy levels.

The passage of 1 mol H2 requires 2 Faradays.

The balanced equation for the electrolysis of water is:
2H2O → 2H2 + O2
For every mole of H2 produced, two moles of electrons are needed to reduce the two protons in each H2O molecule to H2 gas. One Faraday is equal to the amount of electrical charge (i.e., the number of electrons) needed to reduce or oxidize one mole of a substance during an electrolytic reaction. Therefore, for the reduction of one mole of H2O to produce one mole of H2, two Faradays are required.
To find out how many Faradays are required for the passage of 1 mol H2, we'll use the following terms and concepts:
1. Mole (mol): A unit of measurement for the amount of substance, equal to 6.022 x 10^23 particles (Avogadro's number).
2. Hydrogen (H2): A diatomic molecule composed of two hydrogen atoms.
3. Faraday: A unit of electric charge, equal to the charge of 1 mole of electrons, approximately 96,485 Coulombs.
Now let's calculate how many Faradays are needed:
Step 1: Determine the moles of electrons involved in the reaction.
For the formation of H2, the balanced half-reaction is: 2H+ + 2e- → H2
This means that for every 1 mol of H2, 2 moles of electrons (2e-) are involved in the reaction.
Step 2: Calculate the required Faradays.
1 mol of electrons = 1 Faraday (96,485 Coulombs)
So, for 2 moles of electrons, we need 2 Faradays.
Therefore, the passage of 1 mol H2 requires 2 Faradays.

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1 mol H2 requires passage of 2 Faradays.

1. According to Faraday's law of electrolysis, the amount of substance produced or consumed at an electrode during electrolysis is proportional to the charge passed through the cell.
2. For the production of 1 mol H2, we must consider the balanced half-reaction for the reduction of hydrogen ions to form hydrogen gas: 2H+ + 2e- → H2
3. From this half-reaction, we see that 2 moles of electrons (2e-) are required to produce 1 mol of hydrogen gas (H2).
4. Since 1 Faraday is equal to the charge of 1 mole of electrons (approximately 96,485 C/mol), we can conclude that 1 mol H2 requires the passage of 2 Faradays, as 2 moles of electrons are needed for the production of 1 mol H2.

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Answer:

1. Magnesium + Oxygen --> Magnesium oxide

2.Iron + Oxygen --> Iron oxide

3.Copper + Oxygen --> Copper oxide

Explanation:

When an Element such as Magnesium or any other Elements are reacted eith Oxygen it forms an Oxide.

If you ever need the symbol equation here it is too :

1. Mg + O --> MgO

2.Fe + O --> FeO

3.Cu + O--> CuO

Explanation:

1 mole of HNO3 reacts with 1 mole of S to produce 1 mole of H2SO4, 6 moles of NO2, and 2 moles of H2O.

So, when 2 grams of HNO3 (1 mole) reacts, it produces 2 moles of H2O, which is equal to 2 x 18 = 36 grams.

Therefore, 36 grams of water can be made when 2 grams of HNO3 are consumed

Chemical substances that are meant to damage or kill people, animals, or plants are referred to as chemical weapons. Chemical weapons include, for instance:

Nerve agents : which attack the nerve system and can result in respiratory failure and death, include sarin, tabun, VX, and soman.

Blister agents: Agents that produce intense blistering and harm to the skin, eyes, and respiratory system include mustard gas and lewisite.

Blood agents: substances that interfere with the body's capacity to use oxygen and can result in seizures, coma, and death. Examples include hydrogen cyanide and cyanogen chloride.

Choking hazards: These include phosgene and chlorine gas, which can induce fatal respiratory distress.

Consequently, nursing students are able to recognize choking agents, blood agents, blister agents, and nerve agents.

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Answer:

A

Explanation:

The molar mass of Prussian Blue is 860 g / mole

Prussian blue is produced by the oxidation of ferrous  ferrocyanide salts. The molecular formula is represented as follows:

Molecular mass

Molecular mass is a number equal to the sum of the atomic masses of the atoms in a molecule.

Therefore,

molecular mass of  Prussian blue= 56 × 4 + 56 × 3 + 26 × 18

molecular mass of  Prussian blue= 224 + 168 + 468

molecular mass of Prussian blue = 860 g / mole

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It is a second-order reaction since  the order of the reaction with respect to the reactant is 2.

To determine the order of the reaction with respect to the reactant, we'll use the rate law expression and the given information.

Given information: Doubling the concentration of the reactant increases the rate of the reaction by four times. Let's denote the reactant concentration as [A] and the initial rate of the reaction as k. The rate law expression can be written as:

Rate = k[A]^n

Now, we'll use the given information to set up an equation. When the concentration of the reactant is doubled, the rate increases four times:

4 * Rate = k[2A]^n

Now, we can divide this equation by the original rate equation:

(4 * Rate) / (Rate) = (k[2A]^n) / (k[A]^n)

This simplifies to:

4 = (2A)^n / (A)^n

Since A cancels out, we get:

4 = 2^n

To solve for n, we take the base-2 logarithm of both sides:

log2(4) = log2(2^n)

2 = n

So, the order of the reaction with respect to the reactant is 2, meaning it is a second-order reaction.

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Answer:

Explanation:

The pH of a solution can be found by using the formula

\(pH = - log ([ {H}^{+} ])\)

From the question

[H+] = 0.045 M

We have

\(pH = - log(0.045) \\ = 1.346787...\)

We have the final answer as

Hope this helps you

Answer:

1.35

Explanation:

The major products from the reaction of methyl butanoate with diisobutylaluminum hydride at -78 degrees Celsius, followed by acidic work-up, are 2-methylbutanol and isobutyl acetate.

1. Reaction with diisobutylaluminum hydride: Diisobutylaluminum hydride (DIBAL-H) is a strong reducing agent that can convert esters into alcohols. In this case, methyl butanoate undergoes reduction to form 2-methylbutanol.

2. Acidic work-up: After the reduction step, the reaction mixture is treated with an acidic solution. This step helps in the hydrolysis of any remaining DIBAL-H and in the conversion of the intermediate alkoxyaluminum species to the corresponding alcohol and aluminum hydroxide.

Overall reaction:

Methyl butanoate + Diisobutylaluminum hydride → 2-Methylbutanol + Aluminum hydroxide

Additional product: Isobutyl acetate may also be formed as a minor product, resulting from the reaction of diisobutylaluminum hydride with the carbonyl group of the ester.

It is important to note that the reaction conditions, such as temperature and reagent concentrations, can influence the selectivity and yield of the products. The specific reaction conditions used in the experimental setup can provide more detailed information about the major products obtained.

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H+ appear on left hand side and the coefficient is 16 in the balanced redox reaction under acidic aqueous conditions.

What is redox reaction?

Redox reactions, also referred to as oxidation-reduction processes, are reactions in which electrons are transferred from one species to another. An oxidized species is one that has lost electrons, whereas a reduced species has gained electrons.

The balanced redox reaction under acidic aqueous conditions using the smallest whole-number coefficients is written as

\(2MnO+16H^{+} +10I^{-}\) →  \(2Mn2^{+} + 8H_{2} O + 5I^{2}\)

Therefore, the H+ is present on left side with a coefficient of 16 in the balanced redox reaction under acidic aqueous conditions.

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The given solution of magnesium hydroxide is 2L and 0.04M. We will then need 0.8 L of 0.05 M HF to neutralize this solution.

Hydrogen fluoride and magnesium  hydroxide are acid and base respectively and will thus undergo a neutralisation reaction.

The balanced  chemical equation for this  neutralization reaction is:

2HF + Mg(OH)₂ → MgF₂ + 2H₂O

From the equation, youll see  that the  stoichiometric  ratio of HF to Mg(OH)₂ is 2:1. Means, for every 2 moles of HF, 1  mole  of Mg(OH)₂ can be  neutralized.

The given concentration of Mg(OH)₂ is 0.04 M,  and you've got 2 L. Therefore, the number of moles of Mg(OH)₂ in the solution is:

\(n_{Mg(OH)_{2} }\) = M × V = 0.04 mol/L × 2 L = 0.08 mol

To neutralize this amount  of Mg(OH)₂, you'll need half as many moles of HF. so the number of moles of HF that you will need is:

\(n_{HF}\)= 0.08 mol / 2 = 0.04 mol

The concentration of HF is given as 0.05 M, so you'll need:

\(V_{HF}\) = \(n_{HF}\)/ \(M_{HF}\) = 0.04 mol / 0.05 mol/L = 0.8 L of HF

Therefore, you'll need 0.8 L of 0.05 M HF to neutralize 2 L of 0.04 M  Mg(OH)₂.

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Answer:

gas and liquid

Explanation:

right?

The number of the hydrogen atoms that would be required from the diagram is 10.

A Saturated compound has all its carbon atoms connected by single bonds, and each carbon atom is bonded to the maximum number of hydrogen atoms possible. This arrangement allows the compound to have no available or unsaturated bonds for additional atoms.

The compound that is shown must be butane as such the number of the hydrogen atoms that it contains is a total of ten.

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Answer:

the boiling point of HF is higher than H2 and F2

Explanation:

it is because inter molecular give is present in HF because of the Hydrogen bond present. Hydrogen bond is strong and will require much energy to break ,since no hydrogen bond is in the rest H2 and F2 has a low boiling point

The boiling point of a liquid depends upon the pressure of the surrounding. Here HF has higher boiling than the molecule of hydrogen and fluorine due to the presence of hydrogen bonding between 'H' and 'F' atoms.

The temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure of the liquid's environment is defined as the boiling point. A t this temperature, the liquid is converted into a vapour.

The intermolecular forces can be used to predict the relative boiling points. If stronger is the intermolecular force, the lower is the vapor pressure of the substance and the higher is the boiling point.

The order of intermolecular forces are:

H - bonding > dipole-dipole > London dispersion (Van der Waals)

Here HF has intermolecular hydrogen bonding which is not present in H₂ and F₂ and has high boiling point.

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Answer:

For 1 mole of MgCl2, it would require 2 moles of KOH. ( 1 : 2 mole ratio)

Since you have 3 moles of KOH, it is in excess, and MgCl2 is the limiting reactant.

BASIC

bc here:

1,2,3,4,5,6. 7. 8,9,10

The ones in bold are acids.

7 is neutral.

The ones in italic are basic.

To create a buffer solution that maintains a pH of around 7.54, you should choose option (a) CH₃COOH and NaCH₃COO.

A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. It is designed to resist significant pH changes when small amounts of an acid or a base are added.

In this case, CH₃COOH (acetic acid) is a weak acid, and NaCH₃COO (sodium acetate) is its conjugate base. The weak acid and its conjugate base work together to resist pH changes, as the acid can neutralize added base and the conjugate base can neutralize added acid. The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.

The other options are not suitable for creating a buffer with a pH of 7.54. Option (b) HClO and KClO consist of a weak acid and its conjugate base, but their pKa values would not result in the desired pH. Options (c) NaOH and HCN and (d) HNO₃ and KNO₃ consist of a strong base and a weak acid or a strong acid and its conjugate base, respectively, which do not create an effective buffer solution.

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