Calculate the concentration of acetic acid (pka 54.76)and sodium acetate necessary to prepare a 0.2 m buffer solution at pH 5.0

The representative particle for copper metal, Cu, refers to the smallest unit which retains the properties of copper – in this case, a single copper atom.

In chemistry, a representative particle refers to the smallest unit of a substance that still retains the properties of that substance. For a pure metal like copper (Cu), the representative particle is an atom. Therefore, the representative particle for copper metal, Cu, is a copper atom. Copper atoms join together in a crystal lattice structure forming the bulk material we see and use in everyday life like wires, coins etc. But the smallest unit of this structure, the unit which represents its basic physical and chemical properties, is a single copper atom.

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Answer:

22.7%

Explanation:

We must first put down the equation of reaction to guide our work while solving the problem.

KCN(aq) + HCl (aq)--> KCl(aq) + HCN(aq)

The questions specifically state that HCl is in excess so KCN is the limiting reactant.

Number of moles of KCN reacted= mass of KCN reacted / molar mass of KCN

Mass of KCN reacted= 3.2 g

Molar mass of KCN= 65.12 g/mol

Number of moles of KCN= 3.2/65.12 g/mol= 0.049 moles

Theoretical yield of HCN is obtained thus;

From the reaction equation;

1 mol of KCN produced 1 mole of HCN thus 0.049 moles of KCN will produce 0.049 moles of HCN.

Mass of HCN = number of moles ×molar mass

Molar mass of HCN= 27.0253 g/mol

Hence mass of HCN produced= 0.049mol × 27.0253 g/mol= 1.32g of HCN

Actual yield of HCN= 0.30g

% yield= actual yield/ theoretical yield ×100

% yield= 0.30/1.32 ×100

%yield= 22.7%

Answer:

The theoretical yield HCN is 1.33 grams

The percent yield HCN is 22.56 %

Explanation:

Step 1: Data given

Mass of KCN = 3.2 grams

Molar mass KCN = 65.12 g/mol

Mass of HCN produced = 0.30 grams

Step 2: The balanced equation

KCN + HCl --> KCl + HCN

Step 3: Calculate moles KCN

Moles KCN = mass KCN / moalr mass KCN

Moles KCN = 3.2 grams / 65.12 g/mol

Moles KCN = 0.0491 moles

Step 4: Calculate moles HCN

For 1 mol KCN we need 1 mol HCl to produce 1 mol KCl and 1 mol HCN

For 0.0491 moles HCN we'll have 0.0491 moles HCN

Step 5: Calculate mass HCN

Mass HCN = moles HCN * molar mass HCN

Mass HCN = 0.0491 moles * 27.03 g/mol

Mass HCN = 1.33 grams = the theoretical yield

Step 6: Calculate the percent yield HCN

Percent yield = (actual yiel / theoretical yield) * 100 %

Percent yield = (0.30 grams / 1.33 grams)

Percent yield = 22.56 %

Answer: heating the canister

Answer:

heating the canister

Explanation:

Answer:

Percent yield = 50%

Explanation:

Given data:

Mass of CH₄ = 16 g

Mass of O₂ = 32 g

Mass of CO₂ = 11 g

Percent yield of CO₂ = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →  CO₂ + 2H₂O

Number of moles of CH₄:

Number of moles = mass/ molar mass

Number of moles = 16 g /16 g/mol

Number of moles = 1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 32 g /32 g/mol

Number of moles = 1 mol

Now we will compare the moles of CO₂ with both reactant.

                             O₂             :            CO₂

                              2              :               1

                              1               :          1/2×1= 0.5 mol

                          CH₄              :            CO₂

                             1               :               1

Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.

Theoretical yield:

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 11 g/ 22 g × 100

Percent yield = 50%

Answer:

Explanation:

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

Initial (M) 2.00 2.00 0 0

Change (M) −x −x +x +x

Equil (M) 2.00 − x 2.00 − x x x

2 2 c 3

2

c 2

[SO ][NO ]

[SO ][NO]

(2.00 )

=

= −

K

x K

x

Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of

NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.

2.00 − x = 1.30

x = 0.70

Substituting back into the equilibrium constant expression:

2c 2 2c 2

(2.00 )

(0.70)

(2.00 0.70)

= − = −

x KxK

Kc = 0.290

Answer:

Succinate oxidation to fumarate The following reactions transform succinate to regenerate oxalacetate. The first of these reactions is carried out by an oxidation catalyzed by succinate dehydrogenase. The hydrogen acceptor is FAD, since the free energy change is insufficient to allow NAD to interact. The final product is fumarate.

Explanation:

The condensation reaction of GDP + Pi and the hydrolysis of Succinyl-CoA involve the H2O necessary to balance the equation.

Answer:

Excess Vinegar

Explanation:

You will have excess vinegar

Final answer:

The recoil speed of the gun is found using the conservation of momentum. The calculation shows that the gun's recoil speed is approximately 2.4 ms-1. So the correct option is A.

Explanation:

The question posed is related to the concept of conservation of momentum in physics. When a bullet is fired from a gun, the bullet moves forward and the gun is pushed back due to recoil, both objects conserving the system's total momentum. To find the speed of recoil of the gun, you can use the formula:

[tex]m_{bullet}v_{bullet} = m_{gun}v_{gun}[/tex]

Where:

Thus, solving for [tex]v_{gun}[/tex] :

[tex]v_{gun} = m_{bullet}v_{bullet} / m_{gun}[/tex]

[tex]v_{gun}[/tex]= (0.01 kg * 120 m/s) / 0.49 kg = 2.45 m/s

The speed of the recoil of the gun is therefore approximately 2.4 ms-1, which matches answer choice A.

Answer:

1.477mole

Explanation:

First, we'll begin by calculating the molar mass of CO2. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the question = 65g

Mole of CO2 =?

Number of mole = Mass/Molar Mass

Mole of CO2 = 65/44

Mole of CO2 = 1.477mole

Therefore, 65g of carbon dioxide (CO2) contains 1.477mole

Enthalpy change is a measurement of heat energy absorbed or released during a chemical reaction at constant pressure. It reflects the net energy required to break the chemical bonds of reactants versus the energy released when new bonds form in products, determining if the reaction is endothermic or exothermic.

It represents the difference between the enthalpy (heat content) of the reactants and the products. When chemical bonds are broken in reactants, energy is absorbed, and when new bonds form in products, energy is released. If the energy required to break bonds is greater than the energy released when new bonds are formed, the reaction is endothermic and the enthalpy change is positive; conversely, if more energy is released during bond formation than is absorbed during bond breaking, the reaction is exothermic and exhibits a negative enthalpy change.

The enthalpy change can be calculated using the formula ΔH = Σ (bond dissociation energies of reactants) minus Σ (bond dissociation energies of products), which simplifies the overall energy balance of bond breaking and bond formation processes. This understanding allows scientists to predict whether a reaction will absorb or release energy and is crucial in fields such as thermodynamics and chemical engineering.

Answer:First: the coefficients give the number of molecules (or atoms) involved in the reaction. In the example reaction, two molecules of hydrogen react with one molecule of oxygen and produce two molecules of water. Second: the coefficients give the number of moles of each substance involved in the reaction.

Explanation: PLESE GIVE BRAINLIEST

The coefficients tells us about the number of atoms, molecules or compound present in a reaction. They are actually the numbers or terms which are used in a balanced chemical equation.

A chemical equation in which amount of reactants and products on both sides of the equation are equal is defined as the balanced chemical equation. The number of atoms of reactants and products on both sides will be equal.

The coefficients are the numbers which are added in front of the chemical formulae or symbol in order to balance the chemical equation. Only by adding the coefficients a balanced equation obeys the law of conservation of mass.

An equation is balanced with the help of coefficients.

For example, in the given reaction:

2H₂ + O₂ → 2H₂O

The coefficients of H₂,O₂ and H₂O are 2, 1 and 2.

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Answer:

I don't understand

Explanation:

Sorry

Answer:

Butanol

Explanation:

By counting, we know this is C4H9OH

C4H9OH is Butanol

Answer:

Their pH is less than 7

Explanation:

Both of these compounds pHs are greater than 7.

Answer:

A) Their pH is less than 7

Explanation:

Their pH is less than 7 is not correct. Their pH is greater than seven, but not equal to seven.

Answer:

6.56%

Explanation:

mass of solute = 22.8 g

mass of solution = mass (solute) + mass (water) = 22.8 + 325 = 347.8 g

% by mass = [mass(solute)/ mass(solution)]*100% =(22.8/347.8)*100% = 6.56%

Answer:

B. A disease wipes out several plant species.

Explanation:

This will be the most detrimental to the environment as it will decrease oxygen levels and many animals will lose their shelter.

- everything else happens every year and is common and normal.

A disease wiping  out several plant species will most likely cause the sustainability of an ecosystem to weaken.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Answer:

Explanation:

[OH⁻] = 1.07  x 10⁻¹⁰

[H⁺] x [ OH⁻ ] = 10⁻¹⁴

[H⁺] x 1.07  x 10⁻¹⁰ =  10⁻¹⁴

[H⁺] = 10⁻¹⁴ / 1.07  x 10⁻¹⁰

= .93458 x 10⁻⁴

= 9.3458 x 10⁻⁵

= 9.35 x 10⁻⁵

= 9.35e-5. M.

Answer:

0.24 g/mL

Explanation:

The density of an object is given by the ratio between its mass and its volume:

[tex]\rho=\frac{m}{V}[/tex]

where

m is the mass of the object

V is its density

In this problem, we have:

m = 7.0 g is the mass of the object

The volume of an irregular shaped object can be measured by putting it into water, and by measuring the difference in water volume.

In this case,

[tex]V_1=25 mL[/tex] is the initial volume of water

[tex]V_2=54 mL[/tex] is the final volume of water

So the volume of the object is

[tex]V=V_2-V_1=54-25=29 mL[/tex]

Therefore, the density of the object is:

[tex]\rho = \frac{7.0 g}{29 mL}=0.24 g/mL[/tex]

Final answer:

To find the density of an irregular shaped object submerged in water, the volume displaced by the object is calculated first, and then the density is found by dividing the mass of the object by the displaced volume. In this case, the density is 0.241 g/mL.

Explanation:

When an irregular shaped object is placed in a graduated cylinder with a starting volume of 25 mL of water and causes the water level to rise to 54.0 mL upon submersion, the object displaces a volume of water equal to the difference in these measurements. To calculate the density of the object, you have to use the formula for density, which is mass divided by volume. Given that the mass of the object is 7.0 grams and the volume displacement is 54.0 mL - 25 mL = 29.0 mL, the density of the object can be found using these values.

The calculation would be as follows: Density = Mass / Volume = 7.0 grams / 29.0 mL = 0.241 g/mL.

Therefore, the density of the object is 0.241 g/mL, indicating how compact the object's mass is within its volume.

Answer:

Too much phosphorus in the water causes algae to grow faster than ecosystems can handle. Large growths of algae are called algal blooms and they can severely reduce or eliminate oxygen in the water, leading to illnesses in fish and the death of large numbers of fish.

Explanation

Futhermore, with a loss of fish the whole ecosystem will be upset and will be in need of serious recovery.

Final answer:

Phosphate acts as an environmental contaminant by promoting eutrophication, excessive growth of algae in water. It is introduced through human activities and contributes to the overgrowth of algae, leading to negative effects on the aquatic ecosystem.

Explanation:

Phosphate acts as an environmental contaminant due to its role in promoting eutrophication. Eutrophication is the excessive growth of algae in a body of water, which leads to a decrease in dissolved oxygen levels that can harm aquatic organisms. Phosphates are introduced into the environment through human activities, such as the use of phosphate-containing detergents and the leaching of phosphate from rock by weathering. These phosphates contribute to the overgrowth of algae and the subsequent negative effects on the aquatic ecosystem.a

Answer: [NO2][O2]/[NO] [O3]

Explanation: Kc = [NO2][O2]/[NO] [O3]

Answer:

Approximately [tex]\rm 31.5\; g[/tex].

Explanation:

The mass of a solution can be divided into two parts:

In this particular [tex]\rm KOH[/tex] solution in water,

The number [tex]35\%[/tex] here likely refers to the concentration of [tex]\rm KOH[/tex] in this solution. That's ratio between the mass of the solute ([tex]\rm KOH[/tex]) and the mass of the whole solution (mass of solute plus mass of solvent.) That is:

[tex]\displaystyle \frac{m(\text{KOH})}{m(\text{solution})} = 35\% = 0.35[/tex].

Hence, [tex]m(\mathrm{KOH}) = 0.35\, m(\text{solution})[/tex].

However, since the solution contains only the solute and the solvent, [tex]m(\text{solution}) = m(\text{solute}) + m(\text{solvent})[/tex].

For this solution in particular,

[tex]\begin{aligned}&m(\text{solution})\\&= m(\text{solute}) + m(\text{solvent}) \\ &= m(\text{KOH}) + m(\text{water})\end{aligned}[/tex].

As a result,

[tex]\begin{aligned}&m(\mathrm{KOH})\\ &= 0.35\, m(\text{solution}) \\&= 0.35\, (m(\mathrm{KOH}) + m(\text{water}))\\&= 0.35\, m(\mathrm{KOH}) + 0.35 \, m(\text{water})\end{aligned}[/tex].

Subtract [tex]0.35\, m(\mathrm{KOH})[/tex] from both sides of the equation:

[tex](1 - 0.35)\, m(\mathrm{KOH}) = 0.35\, m(\text{water})[/tex].

[tex]\begin{aligned} &m(\mathrm{KOH}) \\ &= \left(\frac{0.35}{1 - 0.35}\right)\cdot m(\text{water}) \\ &= \frac{0.35}{0.65} \times 58.5\; \text{g} = 31.5 \; \text{g}\end{aligned}[/tex].

Note, that for this calculation, there's nothing special about this [tex]35\%[/tex] solution of [tex]\mathrm{KOH}[/tex] in water. In general,

[tex]\displaystyle m(\text{solute}) = \left(\frac{\%\text{concentration}}{100\% - \%\text{concentration}}\right)\cdot m(\text{solvent})[/tex].

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

Answer:

128g of Li, will react in this reaction

Explanation:

Before to start working, we need the reaction:

N₂ and Li react, in order to produce Li₃N (lithium nitride)

N₂ + 6Li → 2Li₃N

1 mol of nitrogen reacts with 6 moles of lithium

We convert the mass of N₂ to moles → 86.1 g . 1 mol/ 28g = 3.075 moles

1 mol of N₂ reacts with 6 mol of Li

Therefore, 3.075 moles of N₂ will react with 18.4 moles of Li

We conver the moles to mass → 18.4 mol . 6.94g / 1mol = 128 g

Answer:

It is not possible, since the total weight of all the reagents involved in the reaction is equal to the total weight of the product, since it is not possible to create or destroy atoms, thus, it is not possible to change the weight of the product. It is equal to the weight of the reagent.

Explanation:

Answer:

32g

Explanation:

We have to remember that for percent (w/w) concentration we usually write;

Percent concentration= mass of solute/mass of solution ×100

Since mass of solute= 14.7 g and percent concentration = 32.2%

Then

Mass of solution= mass of solute × 100/ percent concentration

Mass of solution= 14.7 ×100/32.2

Mass of solution= 46.7 g

Since mass of solution = mass of solute + mass of solvent

Mass of solute= 14.7 g

Mass of solution = 46.7g

Mass of solvent = 46.7g -14.7g = 32g

Answer:

Explanation:

Marine debris consists of most anything used and discarded by humans. Winds, rivers and storms can carry this debris to the ocean from far inland, so it is not solely a case of pollution along shorelines. It is also true that oceangoing ships directly contribute to marine debris while they are at sea and not associated with any particular nation.

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

[tex]P\propto T[/tex]

or,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure = 198 kPa

[tex]P_2[/tex] = final pressure = 225 kPa

[tex]T_1[/tex] = initial temperature = [tex]27^oC=273+27=300K[/tex]

[tex]T_2[/tex] = final temperature = ?

Now put all the given values in the above equation, we get:

[tex]\frac{198kPa}{300K}=\frac{225kPa}{T_2}[/tex]

[tex]T_2=340.9K\approx 341K[/tex]

Therefore, the temperature of the air in the tire is, 341 K

Answer:

0.0144 L

Explanation:

Step 1:

Data obtained from the question. This includes:

Temperature (T) = 298k

Pressure (P) = 1 bar

Time (t) = 1.25 hr

Current (I) = 0.0500 A

Step 2:

Determination of the quantity of electricity (Q) used. This is illustrated below:

Q = it

Time (t) = 1.25 hr = 1.25 x 3600 = 4500 secs

Current (I) = 0.0500 A

Quantity of electricity (Q) =?

Q = it

Q = 0.05 x 4500

Q = 225C

Step 3:

Determination of the number of mole of O2 liberated in the process.

In the electrolytic process, O2 will be liberated according to the equation:

2O^2- + 4e- —> O2

From the above illustration, 4 faraday are needed to liberate 1 mole of O2.

1 faraday = 96500C

Therefore of 4 faraday = 4x96500C = 386000C

From the above equation,

386000C of electricity liberated 1 mole of O2.

Therefore, 225C will liberate = 225/386000 = 5.83x10^-4 mole of O2.

Step 4:

Determination of the volume of the O2 liberated.

Number of mole (n) = 5.83x10^-4 mole

Temperature (T) = 298k

Pressure (P) = 1 bar = 0.987 atm

Gas constant (R) = 0.082atm.L/Kmol

Volume (V) =?

Applying the ideal gas equation:

PV = nRT

The volume of O2 can be obtained as follow:

PV = nRT

0.987 x V = 5.83x10^-4 x298x0.082

Divide both side by 0.987

V = (5.83x10^-4 x298x0.082)/0.987

V = 0.0144 L

The volume of [tex]\(O_2\)[/tex] produced is approximately [tex]0.0145 \ liters[/tex] at [tex]298 \ K[/tex] and [tex]1.00[/tex] bar.

To determine the volume of [tex]\(O_2\)[/tex] gas produced in an electrolytic cell, we can follow these steps:

Calculate the total charge

The total charge [tex]\(Q\)[/tex] passed through the cell is given by the product of current [tex]\(I\)[/tex] and time [tex]\(t\)[/tex]:

[tex]\[ Q = I \cdot t \][/tex]

Given:

[tex]Current, \(I = 0.0500 \, \text{A}\)[/tex]

[tex]Time, \(t = 1.25 \, \text{hr} = 1.25 \times 3600 \, \text{s} = 4500 \, \text{s}\)[/tex]

[tex]\[ Q = 0.0500 \, \text{A} \times 4500 \, \text{s} = 225 \, \text{C} \][/tex]

Determine the amount of [tex]\(O_2\)[/tex] produced

The half-reaction for the production of [tex]\(O_2\)[/tex] in water electrolysis is:

[tex]\[ 2 \, \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \, \text{H}^+ + 4 \, \text{e}^- \][/tex]

From this reaction, we see that [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]

The Faraday constant [tex](\(F\))[/tex] is [tex]\(96485 \, \text{C/mol}\)[/tex].

The moles of electrons [tex](\(n_{\text{e}^-}\))[/tex] can be calculated as:

[tex]\[ n_{\text{e}^-} = \frac{Q}{F} = \frac{225 \, \text{C}}{96485 \, \text{C/mol}} = 0.00233 \, \text{mol} \][/tex]

From the stoichiometry of the half-reaction, [tex]4[/tex] moles of electrons produce [tex]1[/tex] mole of [tex]\(O_2\)[/tex]:

[tex]\[ n_{\text{O}_2} = \frac{n_{\text{e}^-}}{4} = \frac{0.00233 \, \text{mol}}{4} = 0.000583 \, \text{mol} \][/tex]

Calculate the volume of [tex]\(O_2\)[/tex] gas

Use the ideal gas law to find the volume of [tex]\(O_2\)[/tex] gas at [tex]298\ K[/tex] and [tex]1.00[/tex] bar:

[tex]\[ PV = nRT \][/tex]

Given:

Pressure, \[tex](P = 1.00 \, \text{bar} = 1.00 \times 10^5 \, \text{Pa}\)[/tex]

Temperature, [tex]\(T = 298 \, \text{K}\)[/tex]

Gas constant, [tex]\(R = 8.314 \, \text{J/(mol K)}\)[/tex]

Moles of [tex]\(O_2\), \(n = 0.000583 \, \text{mol}\)[/tex]

[tex]\[ V = \frac{nRT}{P} = \frac{0.000583 \, \text{mol} \times 8.314 \, \text{J/(mol K)} \times 298 \, \text{K}}{1.00 \times 10^5 \, \text{Pa}} \][/tex]

[tex]\[ V = \frac{1.447 \, \text{J}}{1.00 \times 10^5 \, \text{Pa}} = 1.447 \times 10^{-5} \, \text{m}^3 \][/tex]

[tex]\[ V = 0.01447 \, \text{L} \][/tex]

Answer:

the answer is option 3.