Consider the balanced equation for the decomposition of ozone:

2O3(g) 3O2(g)

Can the rate law for this chemical reaction be deduced from its balanced equation?

A)Yes; since the reactant and product are both oxygen gases, the reaction takes 0 seconds.

B)Yes; the coefficients of the balanced equation determine the order of the reaction.

C)No; the rate law must be determined experimentally.

D)No; the rate will change randomly.

Answers

Answer 1

Answer:

C)No; the rate law must be determined experimentally.

Explanation:

Rate Laws are empirical relationships defining rates of reaction. For any given reaction The Empirical Rate Law is the product of the concentration of the reactants raised to the power of their order of reaction. That is, for the hypothetical reaction aA + bB => Products its Empirical Rate Law is ...

Rate = k[A]ᵃ[B]ᵇ where k = the rate constant and a & b are orders of reaction.

Order of reaction is a 'rate trend' when one changes a rate factor such as concentration. This can only be determined by experimental observation by physically increasing or decreasing concentration and observing the change in reaction rate relative to a reference reaction of interest. Typically they are whole numbers but can be decimal fractions.  

By graphing experimental outcomes of Rxn Rate vs Change in Concentration one can define the order of reaction. Observations  give ...

0-order reactions => change concentration => no change in rate

1st order reactions => change concentration => proportional linear change in rate

2nd order reactions => change concentration => exponential change in rate.        

     

Consider The Balanced Equation For The Decomposition Of Ozone:2O3(g) 3O2(g)Can The Rate Law For This
Answer 2

Answer:

The answer is C

Explanation:

I had the question on edg


Related Questions

How many moles are in 65 g of carbon dioxide (CO2)?

Answers


The answer is 44.0095. We assume you are converting between grams CO2 and mole.

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

A bullet of mass 10 g is fired from a gun of mass 490 g. The bullet leaves the gun with a speed of 120 ms-1. Find the speed of recoil of the gun.

A. 2.4 ms-1 B. 4.8 ms-1

C. 6.4 ms-1 D. 7.4 ms-1​

Answers

C.6.4 ms-1 that’s the answer

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:

[tex]m_{bullet}[/tex] mass of the bullet (10 g or 0.01 kg)[tex]v_{bullet}[/tex] speed of the bullet (120 m/s)[tex]m_{gun}[/tex]e mass of the gun (490 g or 0.49 kg)[tex]v_{gun}[/tex]the recoil speed of the gun

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.

When an irregular shaped object was put in a graduated cylinder with an initial volume of 25 mL of water, the water level rose to 54.0 mL of water after the object was submerged. If the mass of the object is 7.0 grams, what is the density of the object

Answers

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.

A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was found that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilibrium constant, K, for the following reaction. SO2(g) NO2(g) equilibrium reaction arrow SO3(g) NO(g)

Answers

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

Which part of an atom is most directlly involved in chemical bonding?

Answers

Proton, I have to type a lot so this is that lol

Which situation will most likely cause the sustainability of an ecosystem to weaken?
A zebra is killed and eaten by a lion.
A disease wipes out several plant species.
A frog lays hundreds of eggs that hatch into tadpoles.
A dry season causes a few small trees to die.

Answers

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.

What is an ecosystem?

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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Describe the sources of marine debris and explain why it is a problem with global origins.

Answers

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.

What do the COEFFICIENTS in a chemical reaction represent?

Answers

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.

What is 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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A solution of potassium cyanide, KCN, is made by dissolving 3.2 g of KCN in water. When this solution reacts with excess hydrochloric acid, HCl, it produces 0.30 g of poisonous hydrogen cyanide gas, HCN.
KCN + HCl --> KCl + HCN

(a) Calculate the theoretical yield of HCN.

(b) Calculate the percent yield of HCN.

Answers

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 %

How many liters of o2 at 298 k and 1.00 bar are produced in 1.25 hr in an electrolytic cell operating at a current of 0.0500 a?

Answers

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]

What is the representative particle for copper metal, Cu

Answers

Final answer:

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.

Explanation:

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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A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the percent yield of oxygen in this chemical reaction?Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..

Answers

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 %

86.1 g of nitrogen reacts with lithium, how many grams of lithium will react?

Answers

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

NaOH and bleach have several characteristics in common. They include all BUT one of the characteristics listed.

Answers

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.

You started with only 6.750 g of sodium carbonate, but you ended up with more mass than that at the end of the reaction. According to the Law of Conservation of Mass, matter cannot be created or destroyed, so how is it possible for the product to weigh more than the starting material? (State your answer in 1-2 complete sentences.)

Answers

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:

What is meant by enthalpy change?

Answers

Enthalpy change refers to the overall amount of heat added or lost with each step as you progress through your reaction.

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.


What conditions are required for heat transfer between states of matter to occur? In what
direction will heat transfer occur?

Answers

Key Concepts. The transfer of heat can occur in three ways: conduction, convection, and radiation. Heat transfer occurs between states of matter whenever a temperature difference exists and heat transfer occurs only in the direction of decreasing temperature, meaning from a hot object to a cold object.

Consider the reaction between 15.0 mL of a 1.00 M aqueous solution of AgNO3 and 10.0 mL of a 1.00 M aqueous solution of K2CrO4. When these react, a precipitate is observed. What is present in solution after the reaction is complete

Answers

Final answer:

The two solutions AgNO3 and K2CrO4 react to give a precipitate along with an aqueous solution of KNO3. Hence, after the reaction is complete, KNO3 remains in the solution.

Explanation:

When the two aqueous solutions AgNO3 and K2CrO4 react, they produce solid silver chromate Ag2CrO4 as a precipitate and leave potassium nitrate KNO3 in the solution. The reaction is balanced as 2AgNO3 (aqueous) + K2CrO4 (aqueous) -> Ag2CrO4 (s) + 2KNO3 (aqueous). As the reaction proceeds, silver ions (Ag+) and chromate ions (CrO42-) combine to form the precipitate, leaving potassium ions (K+) and nitrate ions (NO3-) in the solution. Hence, after the reaction, the solution consists of KNO3 because it remains aqueous, with 1.00 M K+ and 1.00 M NO3-.

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Calculate the mass of KOH in a 35% solution that contains 58.5 g of water.

Answers

Answer:

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

Explanation:

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

the solute (the material that was dissolved,) and the solvent.

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

[tex]\rm KOH[/tex] is the solute, whilewater is the solvent.

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].

O3(g)+NO(g)<—>O2(g)+NO2(g) Write the equalibrium expression

Answers

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

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

A gas is sealed in a rigid canister at a temperature of –5.0°C and a pressure of 713 mmHg. Which of the following actions would most likely bring the gas to STP?

heating the canister
removing some of the gas from the canister
transfer some of the gas to a larger canister

Answers

Answer: heating the canister

Answer:

heating the canister

Explanation:

The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure
is 225 kPa. What is the temperature of the air in the tire? Assume that the
volume is constant. Make sure your answer is rounded to nearest whole
number and your final answer has the units of K.*

Answers

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

What percentage of radioactive substance remains after two half-lives

Answers

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

Stoichiometry!

Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.

Answers

Answer:

a) 13.2 moles [tex]2H_{2}O[/tex]

b) 79.33 grams of [tex]2H_{2}O[/tex]

Explanation:

First, we'll need to balance the equation

[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]H_{2}O_{(g)}[/tex]

There are 2 (O) on the left and only one on the right, so we'll add a 2 coefficient to the right.

[tex]H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]

Now there are 4 (H) on the right and only 2 on the left, so we'll add a 2 coefficient to the ([tex]H_{2}[/tex]) on the left.

[tex]2H_{2(g)} + O_{2(g)}[/tex] → [tex]2H_{2}O_{(g)}[/tex]

The equation is now balanced.

a) This can be solved with a simple mole ratio.

4.6 moles [tex]O_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{1 mole O_{2}}[/tex] = 13.2 moles [tex]2H_{2}O[/tex]

b) This problem is solved the same way!

2.2 moles [tex]H_{2}[/tex] × [tex]\frac{2 moles H_{2}O}{2 moles H_{2}}[/tex] = 2.2 moles [tex]2H_{2}O[/tex]

However, this problem wants the mass of [tex]2H_{2}O[/tex], not the moles.

The molecular weight of [tex]2H_{2}O[/tex] is the weight of 4 (H) molecules and 2 (O) molecules (found on the periodic table). So,

4(1.008) + 2(15.999) = 36.03 g/mol

2.2 moles [tex]2H_{2}O[/tex] × [tex]\frac{36.03 g}{1 mol}[/tex] = 79.33 grams of [tex]2H_{2}O[/tex]

Which is the formula mass of Na₂SO₄? *
1. 119 amu
2. 125 amu
3. 142 amu
4. 174 amu

Answers

Answer:

the answer is option 3.

3 is the right answer

Calculate the mass, in grams, of the solvent present in a 32.2% solution that contains 14.7 g of NaBr.

Answers

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

what happens when you mix MORE vinegar than baking soda?

Answers

When vinegar and baking soda are mixed, a new chemical called carbonic acid is made. This carbonic acid immediately decomposes into carbon dioxide gas. When you mix the vinegar and baking soda, it's the carbon dioxide gas that makes the bubbles.

Answer:

Excess Vinegar

Explanation:

You will have excess vinegar

Consider the conversion of succinate to fumarate in the Citric Acid Cycle (reaction below). This reaction is endergonic under standard conditions (ΔGo’≈ 6 kJ/mol). How might this reaction be made favorable under equilibrium conditions? Your answer should include the relationship of this reaction to a canonical electron transport chain (i.e. an electron transport chain that uses oxygen as a terminal electron acceptor).

Answers

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.

What is the hydrogen ion concentration of a substance with a hydroxide ion concentration of 1.07x10-10M?
(Be sure to give your answer in 3 sig figs and the unit. Also type the letter "e" for "X10" just like on webassign
with no spaces)

Answers

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.

Given the half‑reactions and their respective standard reduction potentials 1. Cr 3 + + e − ⟶ Cr 2 + E ∘ 1 = − 0.407 V 2. Cr 2 + + 2 e − ⟶ Cr ( s ) E ∘ 2 = − 0.913 V calculate the standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s).

Answers

Answer:

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V

Explanation:

Here we have

1. Cr³⁺  + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V

2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂  = − 0.913 V

To solve the question, we convert, the E⁰ values to ΔG as follows

ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂

Where:

F = Faraday's constant in calories

n = Number of e⁻

ΔG₁ = Gibbs free energy for the first reaction

ΔG₂ = Gibbs free energy for the second half reaction

E⁰₁  = Reduction potential for the first half reaction

E⁰₂ = Reduction potential for the second half reaction

∴ ΔG₁ = 1 × F × − 0.407 V

ΔG₂ = 2 × F  × − 0.913 V

ΔG₁  + ΔG₂  = F × -2.233 V which gives

ΔG = n × F × ΔE⁰ = F × -2.233 V  

Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons

We have, 3 × F × ΔE⁰ = F × -2.233 V

Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.

Final answer:

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is found by the summation of the standard reduction potentials of the half-reactions involved, giving -1.32V.

Explanation:

To calculate the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s), we first need to understand that a redox reaction is a sum of an oxidation half-reaction and a reduction half-reaction. In the given half-reactions, the first is for Cr3+ being reduced to Cr2+ and the second one is for Cr2+ being reduced to Cr(s).

From the given standard reduction potentials, the 1st reaction has E°1 = -0.407 V and the 2nd reaction has E°2 = -0.913 V. Therefore, to get from Cr3+ to Cr(s), we add both these half-reactions together, which also means we add their potentials together. The sum gives us the potential for the entire reaction, which is E°total = E°1 + E°2 = (-0.407V) + (-0.913V) = -1.32V.

Thus, the standard reduction potential for the reduction half-reaction of Cr(III) to Cr(s) is -1.32V.

Learn more about Reduction Potential here:

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