What percentage of the mass of a carbon-12 atom is
contributedby its electrons? Given the mass of an electron is
(1/1836)amu.

Answer:

The work done and heat absorbed are both -8,1 kJ

Explanation:

The work done in an isobaric process is defined as:

W = -P (Vf - Vi)

Where P is pressure ( 10 atm)

Vf = 10 L

Vi = 2 L

Thus, W = -80 atm×L ≡ -8,1 kJ

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.

I hope it helps!

To prepare the 1.00M aqueous aluminum sulfate working solution, the chemist should pour out approximately 219.78 mL of the aluminum sulfate stock solution.

To prepare a 1.00M aqueous aluminum sulfate working solution, the chemist needs to dilute the 1.82 mol/L aluminum sulfate stock solution. The final volume required is 400 mL.
Using the formula for dilution, (stock concentration) × (stock volume) = (final concentration) × (final volume),
we can rearrange the formula to solve for the stock volume:
Stock volume = (final concentration) × (final volume) / (stock concentration)
Substituting the given values,
Stock volume = (1.00 M) × (400 mL) / (1.82 mol/L) = 219.78 mL

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The chemist should pour out 220 mL of the aluminum sulfate stock solution.

To prepare the desired 400 mL of 1.00M aluminum sulfate solution, the chemist needs to use the concept of dilution, which is described by the equation:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

 where:

-[tex]\( C_1 \)[/tex] is the concentration of the stock solution (1.82 mol/L),

- [tex]\( V_1 \)[/tex] is the volume of the stock solution needed (unknown),

- [tex]\( C_2 \)[/tex] is the concentration of the working solution (1.00M),

-[tex]\( V_2 \)[/tex] is the volume of the working solution (400 mL).

 First, we need to ensure that all volumes are in the same unit, so we'll convert 400 mL to liters:

[tex]\[ 400 \text{ mL} = 0.400 \text{ L} \][/tex]

 Now we can plug the known values into the dilution equation:

[tex]\[ (1.82 \text{ mol/L})V_1 = (1.00 \text{ mol/L})(0.400 \text{ L}) \][/tex]

Solving for [tex]\( V_1 \):[/tex]

[tex]\[ V_1 = \frac{(1.00 \text{ mol/L})(0.400 \text{ L})}{1.82 \text{ mol/L}} \][/tex]

 [tex]\[ V_1 = \frac{0.400}{1.82} \text{ L} \][/tex]

[tex]\[ V_1 \approx 0.220 \text{ L} \][/tex]

Converting liters back to milliliters:

[tex]\[ 0.220 \text{ L} = 220 \text{ mL} \][/tex]

 Therefore, the chemist should pour out 220 mL of the 1.82 mol/L aluminum sulfate stock solution to make 400 mL of a 1.00M solution.

Answer : The de Broglie wavelength will be [tex]3.68\times 10^{-33}m[/tex]

Solution :

The formula used for de Broglie wavelength is:

[tex]\lambda=\frac{h}{mv}[/tex] ..........(1)

where,

[tex]\lambda[/tex] = wavelength  = ?

h = Planck's constant  = [tex]6.626\times 10^{-34}Js[/tex]

m = mass  = 1.2 g = 0.0012 kg

Conversion used : 1 kg = 1000 g

v = velocity  = 150 m/s

Now put all the given values in equation 1, we get:

[tex]\lambda=\frac{6.626\times 10^{-34}Js}{(0.0012kg)\times (150m/s)}[/tex]

[tex]\lambda=3.68\times 10^{-33}m[/tex]

Therefore, the de Broglie wavelength will be [tex]3.68\times 10^{-33}m[/tex]

Answer:

The molarity of HCl is 0.138 M

Explanation:

The titration reaction is as follows:

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:

At the endpoint:

moles Ca(OH)₂  = moles HCl / 2

Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:

mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂

mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂

The number of moles of HCl will be:

mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl

This number of moles was present in 50.0 ml, then, in 1000 ml:

mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol

Then:

Molarity HCl = 0.138 M

Answer:

Rate law: = [tex]k[CS_2]^1[/tex]

Explanation:

Given:

t               [tex][CS_2][/tex]

0.100            [tex]2.7 \times 10^{-7}[/tex]

0.080          [tex]2.2 \times 10^{-7}[/tex]

0.055         [tex]1.5\times 10^{-7}[/tex]

0.044         [tex]1.2\times 10^{-7}[/tex]

Rate law for the given reaction: [tex]k[CS_2]^n[/tex]

Where, n is the order of the reaction.

Divide rate 1 with rate 3

[tex]\frac{0.100}{0.055} =\frac{k[CS_2 (1)]^n}{k[CS_2 (3)]^n} \\\frac{0.100}{0.055} =\frac{k[2.7 \times 10^{-7}]^n}{k[1.5\times 10^{-7}]^n}\\1.81=[1.8]^n\\ n=1[/tex]

So, rate law = [tex]k[CS_2]^1[/tex]

Answer:

[tex]r=-3.73x10^{5}s^{-1} [CS_2][/tex]

Explanation:

Hello,

In this case, a linealization helps to choose the rate law for the decomposition of CS₂ as it is generalized via:

[tex]r=-k[CS_2]^{n}[/tex]

Whereas [tex]n[/tex] accounts for the order of reaction, which could be computed by linealizing the given data using the following procedure:

[tex]-ln(r)=ln(k[CS_2]^{n})\\-ln(r)=ln(k)+ln([CS_2]^{n})\\-ln(r)=ln(k)+n*ln([CS_2])[/tex]

Therefore, on the attached picture you will find the graph and the lineal equation wherein the slope is the order of the reaction and the y-axis intercept the natural logarithm of the rate constant. In such a way, the order of reaction is 1 and the rate constant is:

[tex]ln(k)=12.83\\k=exp(12.83)\\k=3.73x10^{5}s^{-1}[/tex]

Best regards.

Answer:

The concentration of the solution in molarity is 4.45*10^-6M

The concentration of the solution in ppm is 1.88

Explanation:

First of all we have to find out how many mols are 0.000333 g of fluorescein. We would use the molar mass of it.

So 0.000333g / 332.32 g/m = 1.002*10^-6 moles

This moles are in 225 ml of ethanol. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)

225 ml ______ 1.002*10^-6 moles

1000 ml _____ x    x = (1000*1.002*10^-6) /225 = 4.45*10^-6M

The ppm means mg of solute that are in 1 kg of solvent.

Let's consider density of ethanol to get the mass.

d = m/v

0.785 g/ml = m/225 ml

0.785 g/ml . 225 ml = m

176.625 g = m

Now, let's convert

0.000333 g *1000 = 0.333 mg

176.625g / 1000 = 0.176625 kg

0.333/0.176625 = 1.88 ppm

Answer:

- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers".

- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential.

Explanation:

Hello, here the answers:

- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers" (taken from Kundu, P. K., Cohen, I. M., & Dowling, D. R. (2012). Fluid mechanics.), thus, it means that when you have a fluid with an acting-on-it share stress (an external force which move the fluid), the related velocity gradient (variation or change in velocity) at which the layers are moving are related as:

[tex]\pi =-v \frac{du}{dy}[/tex]

Whereas [tex]\pi[/tex] is the shear stress, [tex]v[/tex] is the viscosity and the differential accounts for the change in the velocity in the arbitrary [tex]y[/tex] coordinate.

- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential, based on:

[tex]\pi =-v (\frac{du}{dy})^n[/tex]

Whereas [tex]n[/tex] accounts for a decreasing or increasing behavior of the shear stress.

Best regards.

Explanation:

Molecular formulas show correct and accurate number of each type of the atoms which are present in molecule.

On the other hand, structural formulas show arrangement of atoms and covalent bonds between them.

For example,

The molecular formula for carbon dioxide is [tex]CO_2[/tex]

The structural formula is O = C = O

A molecular formula gives the simplest whole number ratio of atoms in a compound, while a structural formula shows the actual arrangement of atoms and bonding in a molecule.

A molecular formula is a way to represent a compound using the symbols of the elements present and the number of atoms of each element in the compound. It gives the simplest whole number ratio of atoms in the compound. For example, the molecular formula of glucose is C6H12O6, indicating that it contains 6 carbon, 12 hydrogen, and 6 oxygen atoms.

A structural formula shows how the atoms in a molecule are connected to each other. It represents the actual arrangement of the atoms in the compound, including the bonding between them. For example, the structural formula of glucose shows the specific arrangement of carbon, hydrogen, and oxygen atoms in the molecule.

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Answer:

We will need 147.772 mL of KH2PO4 to make this solution

Explanation:

For this case we can give the following equation:

H2PO4 - ⇄ H+ + HPO42-

With following pH- equation:

pH = pKa + log [HPO42-]/[H2PO4-]

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

-0.16 =  log [HPO42-]/[H2PO4-]

10^-0.16 = [HPO42-]/[H2PO4-]

0.6918 = [HPO42-]/[H2PO4-]

Let's say the volume of HPO42-= x  then the volume of H2PO4- will be 250 mL - x

Since both have a concentration of 1M = 1 mol /L

If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

0.6918 = x / (250 - x)

0.6918*250 - 0.6918x = x

172.95 = 1.6918x

x = 102.228 mL

The volume of HPO42- = 102.228 mL

Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL

To control this we can plug this in the pH equation

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

7.05= 7.21 + log (102.228 / 147.772) = 7.05

We will need 147.772 mL of KH2PO4 to make this solution

To produce the buffer solution, use the Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid, ensuring that the concentration of KH2PO4 is higher than K2HPO4, and calculate the necessary volumes of 1.00 M stock solutions.

To prepare a potassium dihydrogen phosphate buffer solution with a pH of 7.05, when the pKa of H2PO4− is 7.21, the Henderson-Hasselbalch equation can be used, which is pH = pKa + log([A−]/[HA]), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. Since the desired pH is slightly less than the pKa, you'll need a bit more acid (KH2PO4) than the base (K2HPO4). To find the exact amounts, set up the equation like this: 7.05 = 7.21 + log([K2HPO4]/[KH2PO4]). Solving for the ratio [K2HPO4]/[KH2PO4], we get a value less than 1, which indicates that the concentration of KH2PO4 must be higher than that of K2HPO4 in our buffer solution. Then, you can calculate the volume of each stock solution needed to reach the final total volume of 250 mL, remembering the molarities are both 1.00 M and that volumes are additive.

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Answer:

a) 44.442 grams of aluminium chloride is produced.

b) 8.996 grams of the excess reactant that id aluminum is left unreacted.

Explanation:

[tex]2Al(s) + 3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

Moles of aluminum = [tex]\frac{13.49 g}{27 g/mol}=0.4996 mol[/tex]

Moles of chlorine gas = [tex]\frac{35.45}{35.5 g/mol}=0.4993 mol[/tex]

According to reaction, 1 mol aluminum reacts with 3 moles of chlorine gas.

Then 0.4996 moles of aluminum will react with:

[tex]\frac{3}{1}\times 0.4996 mol=1.4988 mol[/tex] of chlorine gas

Then 0.4993 moles of chlorine gas will react with:

[tex]\frac{1}{3}\times 0.4993 mol=0.1664 mol[/tex] of aluminum

As we can see that moles of an aluminium are in excess. Hence, aluminium is an excess reagent.

So moles of aluminum chloride will depend upon moles of chlorine gas.

According to reaction 3 moles of chlorine gas give 2 moles of aluminium chloride.

Then 0.4993 mole of chlorine gas will give:

[tex]\frac{2}{3}\times 0.4993 mol=0.3329 mol[/tex] of aluminium chloride .

Mass of 0.3329 moles of aluminium chloride :

=0.3329 mol × 133.5 g/mol= 44.442 g

44.442 grams of aluminium chloride is produced.

Moles of aluminium left unreacted = 0.4996 mol - 0.1664 mol = 0.3332 mol

Mass of 0.3332 moles of aluminum :

0.3332 mol × 27 g/mol = 8.996 g[/tex]

8.996 grams of the excess reactant that id aluminum is left unreacted.

Using stoichiometry, we calculate how much aluminum chloride is formed from given masses of aluminum and chlorine and determine the excess reactant remaining. This requires finding the limiting reactant, using the balanced chemical equation, and converting between moles and grams.

To answer the question about how much aluminum chloride is produced when 13.49 g of Al and 35.45 g of Cl2 react, we need to use stoichiometry, which is the method of quantitatively relating the products and reactants in a chemical reaction. Here is the step-by-step explanation:

To find the excess reactant remaining:

Answer:

b. 4.4

Explanation:

pH is related to the concentration of H₃O⁺ through the following equation:

pH = -log([H₃O⁺]) = -log(4 x 10⁻⁵)

pH = 4.4

Answer:

Work is given by -PdV.

It clearly means that when volume of the system changes, work is done either on the system or by the system depending upon volume change.

So, pressure change term -VdP is not included because it is indicating a constant volume.

But in case of enthalpy,

 H = U + PV

So, for enthalpy change we can write

dH = dU + PdV + VdP

Because enthalpy change is the sum of all energy changes in the system that includes internal energy, P-V work and effect of pressure change.

Final answer:

The expression for work in thermodynamics is typically shown as -pdV to denote work done by the system, and is path-dependent, while the enthalpy change includes PdV and VdP terms to represent heat and work interactions at constant pressure, since enthalpy is a state function.

Explanation:

The question relates to why the expression for work in thermodynamics is given as work done by the system (-pdV) rather than the work done on the system, and why in the expression for enthalpy change (dH) both PdV and VdP terms appear. In thermodynamics, work done by a system during expansion is often expressed as W = PextΔV, where Pext is the external pressure and ΔV is the change in volume. The negative sign in -pdV signifies that the system does work on its surroundings, thus losing energy. This convention is such that energy leaving the system is negative, aligning with the conventional definition of heat flow being negative when it leaves the system.

Enthalpy (H) is a state function, which unlike work or heat, is not path-dependent. The differential change in enthalpy, dH, for a small, reversible process is given by dH = dU + PdV + VdP, involving both pressure-volume work and the work done on the system due to pressure changes at constant volume. This dH formula accurately describes the total heat and work interactions of a system at constant pressure. The inclusion of VdP in the expression accounts for the work involved when there is a change in pressure at constant volume.

It is also important to note that the exact expression for work can vary depending on the specific thermodynamic process, such as isothermal (PV=nRT) or adiabatic processes and can become complex when accounting for non-ideal behaviors of gases, hence the significance of defining a when and b to zero for an ideal gas equation.

Answer:

13.4g of NaCl will crystallize per 100g of water at 40°C.

Explanation:

Solubility stands for the maximum amount of solute that may stay dissolved per 100 g of water at a specific temperature. A solution with this concentration is known as a saturated solution. At 40°C, every 100 g of water, 36.6 g of NaCl can remain dissolved. If more solute is added to the solution the excess will crystallize and precipitate. Since prior to criystallization there are 50 g of NaCl dissolved, what will precipitate will be:

50g - 36.6g = 13.4g

Once crystallization is initiated in a 50% sodium chloride solution at 40°C, 13.4 g of NaCl crystals will form per 100 g of water. This is calculated by subtracting the solubility limit at 40°C (36.6 g) from the amount initially present in the solution (50 g).

If a sodium chloride (NaCl) solution in water at 40°C has a concentration of 50%, for every 100 g of water, there is 50 g of NaCl dissolved. Since the solubility of NaCl at 40°C is 36.6 g per 100 g of water, the solution is supersaturated because it contains more solute than can remain in solution at that temperature. When crystallization starts, the excess NaCl above its solubility limit will precipitate out.

To calculate the quantity of NaCl crystals that will form, we subtract the maximum solubility at 40°C from the total amount of NaCl currently dissolved:

Amount of NaCl present initially = 50 g

Maximum solubility at 40°C = 36.6 g

NaCl crystals that will form = Amount of NaCl present initially - Maximum solubility at 40°C

NaCl crystals that will form = 50 g - 36.6 g = 13.4 g

Therefore, 13.4 g of NaCl crystals will form per 100 g of water once crystallization has been started.

Explanation:

The hybridization of the carbon in which it forms 120° with the other atoms is [tex]sp^2[/tex].

In [tex]sp^2[/tex] hybridized carbon, it forms a geometry of trigonal planar which has angle of 120°. This corresponds to doubly bonded carbon.

To draw:

One carbon-carbon double bond in 4 membered carbon chain.

The structure is shown in image below.

Answer:

2300j

Explanation:

Answer:

Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed. [tex]H_2+\frac{1}{2}O_2\rightarrow H_2O[/tex]

Example: 1. reactivity of substances

2. Toxicity of substances

Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc. [tex]H_2O(s)\rightarrow H_2O(l)[/tex]

Example: 1. Melting Point

2. Density

Final answer:

Physical properties describe the characteristics of matter that can be observed or measured without changing its identity, while chemical properties describe how matter changes its chemical structure or composition.

Explanation:

Physical properties are characteristics that describe matter, such as size, shape, color, and mass, that can be observed or measured without changing the identity of the matter. Examples of physical properties include the melting point of ice and the boiling point of water. Chemical properties, on the other hand, describe how matter changes its chemical structure or composition. Examples of chemical properties include flammability, reactivity, and the ability to undergo oxidation. For instance, the ability of gasoline to burn is a chemical property.

Explanation:

A. 100°C to Kelvins

[tex]T(K)=T(^oC)+273.15[/tex]

[tex]T(K)=100(^oC)+273.15=373.15 K[/tex]

B 600°R to Kelvins

[tex](T)^oK=((T)^oR)\times 1.8[/tex]

[tex](T)^oK=600\times 1.8 K = 1080 K[/tex]

C. 98°F to Kelvins

[tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]

[tex](T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K[/tex]

D. 77.4°F to degree Celsius

[tex]((T)^oC)=((T)^oF-32)\times \frac{5}{9}[/tex]

[tex](T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC[/tex]

E. 77.4 K to degree Celsius

[tex]T(^oC)=T(^K)-273.15[/tex]

[tex]T(^oC)=77.4(K)-273.15=-195.75^oC[/tex]

F. 77.4°R to degree Celsius

[tex](T)^oC=((T)^oR-491.67)\times \frac{5}{9}[/tex]

[tex](T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC[/tex]

Answer:

0,040 M

Explanation:

The global reaction of the problem is:

Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40

The equation of equilibrium is:

K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]

The concentration of OH⁻ is:

pOH = 14 - pH = 3

pOH = -log [OH⁻]

[OH⁻] = 1x10⁻³

Thus:

40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]

0,04M =  [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]

This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.

I hope it helps!

Explanation:

The pulp and the paper mill industry is intensive consumer of the water and the natural resources like wood and on return, discharging an ample variety of the gaseous, liquid and solid wastes to environment.

The wastewater which is being eliminated from pulp and paper industry usually contains high and tremendous levels of concentrations of the biochemical oxygen demand and the chemical oxygen demand and thus, shows high level of toxicity and a strong and deep black-brown color.

Answer:

c. C-F.

Explanation:

Hello, to know the type of bond, you must subtract the electronegativities for each element forming the bond:

a. C-N=3.04-2.55=0.49

b. C-C=2.55-2.55=0

c. C-F=3.98-2.55=1.43

d. C-H=2.55-2.2=0.35

Thus, since for an electronegativity difference between 0.7 and 1.7, the bond is polar, the highest number (most polar), based on calculations is 1.43, that is for c. C-F.

Best regards.

Answer:

Rate of reaction is [tex]0.0950M.s^{-1}[/tex]

Explanation:

Answer:

a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population.  ( t = 5.489 E-13 day )

b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population

Explanation:

Let P = 7.7 billion people = 7.7 E12 person

∴ water needed for the total population for one day:

⇒ water amount  = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day

⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year

∴ freshwater available:

freshwater = 6.34 E2 Km² H2O

how long will this water sustain the current population?

⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day

this amount of freshwater does not even meet the need of the current global population.

∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature

⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )

⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year

⇒ Py > water amount

the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.

The average rate of decomposition of NOBr over the entire experiment is -6.7 x 10^(-4) mol/(L*s), and the average rate of decomposition between 2.00 and 4.00 seconds is -8.0 x 10^(-4) mol/(L*s).

The average rate of decomposition of NOBr over the entire experiment can be determined by calculating the change in concentration of NOBr over the change in time. From the given data, the initial concentration of NOBr is 0.0100 mol/L and the final concentration is 0.0033 mol/L after 10 seconds. Therefore, the average rate of decomposition is (0.0033 mol/L - 0.0100 mol/L) / (10 s - 0 s) = -6.7 x 10^(-4) mol/(L×s).

The average rate of decomposition of NOBr between 2.00 and 4.00 seconds can be calculated using the same method. The initial concentration at t=2.00 s is 0.0071 mol/L and the concentration at t=4.00 s is 0.0055 mol/L. Therefore, the average rate of decomposition between 2.00 and 4.00 seconds is (0.0055 mol/L - 0.0071 mol/L) / (4.00 s - 2.00 s) = -8.0 x 10^(-4) mol/(L×s).

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The average rate of decomposition of NOBr over the entire experiment is 0.00067 M/s, while the average rate between 2.00 and 4.00 seconds is 0.0008 M/s.

(a) the average rate of decomposition of NOBr over the entire experiment

We need to calculate the average rate of decomposition of NOBr over the entire given time period. The formula to calculate the average rate of decomposition is:

Rate = - Δ[NOBr] / Δt

Where Δ[NOBr] is the change in concentration and Δt is the change in time.

Initial concentration, [NOBr]0 = 0.0100 M
Final concentration, [NOBr] = 0.0033 M
Initial time = 0.00 s
Final time = 10.00 s

Δ[NOBr] = [NOBr] - [NOBr]0 = 0.0033 M - 0.0100 M = -0.0067 M
Δt = 10.00 s - 0.00 s = 10.00 s

Average Rate = - (-0.0067 M) / (10.00 s) = 0.00067 M/s

(b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds

Initial concentration, [NOBr]2.00s = 0.0071 M
Final concentration, [NOBr]4.00s = 0.0055 M
Initial time = 2.00 s
Final time = 4.00 s

Δ[NOBr] = [NOBr]4.00s - [NOBr]2.00s = 0.0055 M - 0.0071 M = -0.0016 M
Δt = 4.00 s - 2.00 s = 2.00 s

Average Rate = - (-0.0016 M) / (2.00 s) = 0.0008 M/s

Answer:

[tex]\large \boxed{\textbf{1.48 mol/L}}[/tex]

Explanation:

We must use the Nernst equation

[tex]E = E^{\circ} - \dfrac{RT}{zF}lnQ[/tex]

1. Calculate E°

Anode:     Pd²⁺ (0.498 mol·L⁻¹) + 2e⁻ ⇌ Pd;                               E° = +0.987 V

Cathode: Cu ⇌ Cu⁺ (x mol·L⁻¹) + e⁻;                                          E°=  - 0.521 V

Overall:   Pd²⁺(0.498 mol·L⁻¹) + 2Cu ⟶ Pd + 2Cu⁺ (x mol·L⁻¹); E° =   0.466 V

2. Calculate Q

[tex]\begin{array}{rcl}0.447 & = & 0.466 - \dfrac{8.314\times 298}{2 \times 96 485} \ln Q\\\\-0.019& = & -0.01284 \ln Q\\\ln Q & = & 1.480\\Q & = & e^{1.480}\\ & = & 4.392\\\end{array}[/tex]

3. Calculate [Cu⁺]

[tex]\begin{array}{rcl}Q & = & \dfrac{\text{[Cu$^{+}$]}^{2}}{\text{[Pd]}}\\\\4.392 & = & \dfrac{{x}^{2}}{0.498}\\\\x^{2}& = & 2.187\\x & = & 1.48\\\end{array}\\\text{The concentration of Cu$^{+}$ is $\large \boxed{\textbf{1.48 mol/L}}$}[/tex]

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

[tex]AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}[/tex]

Hence all three statements are true

Answer:

1.7927 mL

Explanation:

The mass of solid taken = 4.75 g

This solid contains 21.6 wt% [tex]Ba(NO_3)_2[/tex], thus,

Mass of [tex]Ba(NO_3)_2[/tex] = [tex]\frac {21.6}{100}\times 4.75\ g[/tex] = 1.026 g

Molar mass of [tex]Ba(NO_3)_2[/tex] = 261.337 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{1.026\ g}{261.337\ g/mol}[/tex]

[tex]Moles= 0.003926\ mol[/tex]

Considering the reaction as:

[tex]Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3[/tex]

1 moles of [tex]Ba(NO_3)_2[/tex] react with 1 mole of [tex]H_2SO_4[/tex]

Thus,

0.003926 mole of [tex]Ba(NO_3)_2[/tex] react with 0.003926 mole of [tex]H_2SO_4[/tex]

Moles of [tex]H_2SO_4[/tex] = 0.003926 mole

Also, considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

Molarity = 2.19 M

So,

[tex]2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}[/tex]

Volume = 0.0017927 L

Also, 1 L = 1000 mL

So, volume = 1.7927 mL

Answer

Take account the molar mass of this compound (first of all u should know the formula, S2F10) so when u have the molar mass, you can get how many grams of it, are present in 2.15 moles. The right mass is 546,3 g.

Explanation:

Molar mass in the S2F10 should be, molar mass in S (32.06 x2) + molar mass of F (18.998 x 10) = 254,1 g/m

Answer:

50%

Explanation:

The percent s - character , in a particular hybridization can be calculated from the following formula , i.e. ,

% s character = 1 / ( 1 + x ) * 100

Where , x = number of p orbitals .

Hence , from the question ,

The % s character in sp hybrid orbital is calculated as -

% s character = 1 / ( 1 + x ) * 100

x = 1 ( since , one p orbital is present in the sp hybridization )

% s character = 1 / ( 1 + 1 ) * 100

% s character = 1/ 2 * 100

% s character = 50 %.

Hence , the % s character in  sp hybrid orbital = 50 % .

In sp hybridization, one s and one p atomic orbital combine to form two new hybrid orbitals. Each of these sp hybrid orbitals has 50% s character and 50% p character, so the percent s character in a sp hybridized orbital is 50%. This kind of hybridization is fundamental to understanding molecular geometry in chemistry.

The term sp hybridization refers to the mathematical combination of one s and one p atomic orbital to form two new hybrid orbitals. Each hybrid orbital in a sp configuration contains 50% s character and 50% p character, therefore, the percent s character in a sp hybridized orbital is 50%. This orbital hybridization concept is a fundamental aspect of understanding molecular geometry in chemistry.

As an example, consider carbon. In its ground state, carbon has two unpaired electrons in separate 2p orbitals. If one of its 2s electrons is excited to a 2p orbital, the atom can then hybridize these three orbitals (one 2s orbital and two 2p orbitals) to form an sp hybridized state. The result of this is two equivalent sp orbitals arranged linearly at 180° to each other. This sp hybridization in carbon is seen in molecules like carbon dioxide (CO2).

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Answer:

202.20 cm^3/s

Explanation:

Hello,

I'm sending a photo showing the solution for this exercise.

The solution of the 2x2 system could be achieved by using a calculator or any  available algebraic method.

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