What is the percent hydrolysis in 0.075 M sodium acetate, NaCH3COO, solution?
a. 0.0087%
b. 0.012%
c. 0.0064%
d. 0.0038%
e. 0.043%

During evaporation, for the water molecules to escape as gas, it needs to break hydrogen bonding between molecules.

Various special properties exhibited by water is due to the presence of hydrogen bonding.  When in liquid form, hydrogen bonds are continuously formed and breaks when the molecules moves. Hydrogen bonds breaks as a result of kinetic energy which makes the molecules move in the system. As the temperature is increased kinetic energy increases and these hydrogen bonds are completely broken and escapes as water vapor.

It requires as much heat to break the hydrogen bonds, as much as increasing the temperature. Due to hydrogen bonds water has high specific heat capacity and also high heat of vaporization.

So to vaporize, hydrogen bonds between water molecules must be broken.

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Answer:

c. the reactants

Explanation:

the reactants are specifically the substances consumed in the course of chemical reaction

Ni(OH)₂ is base but this is not strong base this is very weak base and this cannot gives OH⁻ after dissolving in H₂​​​O . this is only dissolve in acidic solution.

A solution is said to be acidic if it contains more positively charged hydrogen ions (H⁺) than negatively charged hydroxide ions (OH⁻). Lemon juice and vinegar are typical examples of acids. The amount of OH ions is greater in bases. Baking soda and regular ammonia are common examples.

Acidic substances can typically be distinguished by their sour flavor. A basic definition of an acid is a molecule that can donate an H⁺ ion and can continue to be energetically favorable even after losing H⁺. Blue litmus paper has a reputation for turning red when exposed to acids. Bases, on the other hand, are distinguished by their bitter flavor and slick consistency.

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We intend to prepare 500 ml of a 0.1m solution of NaCl, but you forget to remove the stir bar when we bring the solution to volume. the stir bar has a volume of 1 ml and weighs 4 g, the percent error in our molarity is 0.2%.

Percent errors indicate how big our errors are when we measure something in analysis process. Smaller percent errors indicate that we are close to accepted or original value.

For example, a 1% error indicates that we got very close to accepted value, while 48% means that we were quite a long way off from the true value. Measurement errors are often unavoidable due to certain reasons like hands can shake, material can be imprecise, or our instruments just might not have capability to estimate exactly.

The percent error formula will let us know how seriously these inevitable errors influenced our results.

Percent error is difference between estimated value and the actual value in comparison to the actual value and is expressed as a percentage. In other words, the percent error is relative error multiplied by 100.

The percent error formula

Formula for percent error is:

PE = (Estimated value-Actual value|/ The Actual value) × 100

0.1% M NaCl

Amount of NaCl require to prepare 0.1M NaCl calculated:

M1= \(\frac{1000*x}{58.5*500}\)

x= 2.95g

but stir bar consume 1ml so actual value of solution is not 500ml but it is 500-1= 499ml

so, molarity= \(\frac{1000*2.925}{58.5*499}\)

M2= 0.1002M

%error= [(M1-M2)/M1]*100

          =0.2%

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Answer:

Hello There!!

Explanation:

The answer is C: evaporation because evaporation is the process in which a liquid boils then evaporates and lastly becomes a gas.

hope this helps,have a great day!!

~Pinky~

According to the forces of attraction, when you boil a pot of liquid water, the water turns to steam and rises in the air. This is an example of evaporation.

Forces of attraction  is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different  states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases  as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.

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a.bond energy

c. endothermic reaction

d. exothermic reaction

f.exothermic

g.endothermic

According to the graph, the concentration of A decreases with time before leveling out. Option A.

The reaction shown is that of a reversible reaction in which A is on the reactant's side and A2 is on the product's side.

At the beginning of the reaction, the concentration of A decreases as a result of forming A2. In other words, the concentration of A2 increases just as that of A decreases.

With time, the reaction reaches an equilibrium during which the rate of formation of A equals the rate of formation of A2. At this point, the concentration of A levels off.

In summary, the concentration of A first decreases before leveling off.

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Answer:

Is C

Explanation:

Answer: It will change to gas

Explanation: It will change to gas because of evaporation. When the heat intensifies it will happen.

The conversions are 2000 mg to g id 2 g, 5 ml to l is 0.005 l, 104 km to m is 104000 m, 198 g to kg is 0.198 kg, 2500 m to km is 2.5 km, 75 ml to l is 0.075 l, 65 g to mg is 65000 mg, 5.6 kg to g is 5600 g,8 mm to dm is  0.0008 dm, 5.6 m to cm is 560 cm and 120 dag to dg is  12000 dg.

Conversion is defined as a client action that is performed on your website and is tracked by conversion rate.

1 mg = 0.001 g

1 L = 1000 ml

1 km = 1000 m

1 g = 0.001 kg

1 mm = 0.0001 dm

1 m = 100 cm

1 dag = 100 dg

Thus, the conversions are 2000 mg to g id 2 g, 5 ml to l is 0.005 l, 104 km to m is 104000 m, 198 g to kg is 0.198 kg, 2500 m to km is 2.5 km, 75 ml to l is 0.075 l, 65 g to mg is 65000 mg, 5.6 kg to g is 5600 g,8 mm to dm is  0.0008 dm, 5.6 m to cm is 560 cm and 120 dag to dg is  12000 dg.

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Answer:

982.5 kg/m³

Explanation:

When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:

ρ₁ = ρ₀/(1 + β*(t₁ - t₀))

Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.

At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C

ρ₁ = 1,000/(1 + 0.0002*(93 - 4))

ρ₁ = 1,000/(1+ 0.0178)

ρ₁ = 982.5 kg/m³

Answer:

Paper Chromatography

Explanation:

Most leaves are green due to chlorophyll pigments of different types like chlorophyll A, chlorophyll B, etc. All these pigments will be seperated according to their molecular composition. So, the correct option is 'Paper Chromatography'.

Answer:

1. Amine.

2. Alcohol.

3. Carboxylic Acid.

4. Halocarbon.

Explanation:

The correct answer according to the tile are Amine, Alcohol, Carboxylic acid, Halocarbon.

How can hydrocarbons be classified based on their structure?

Hydrocarbons can be classified as either aromatic or aliphatic compounds, depending on the presence of a benzene ring.

What is the most common classification of hydrocarbons?

Alkanes are hydrocarbons in which all of the bonds are single bonds. Alkenes are hydrocarbons that contain a carbon-carbon double bond.

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Answer:

3.98dm³

Explanation:

Using combined gas law equation:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (dm³)

V2 = final volume (dm³)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in this question:

V1 = 3.5dm³

V2 = ?

P1 = 101 kPa

P2 = 85.0 kPa

T1 = 23.0°C = 23 + 273 = 296K

T2 = 10.0°C = 10 + 273 = 283K

Using P1V1/T1 = P2V2/T2

101 × 3.5/296 = 85 × V2/283

353.5/296 = 85V2/283

Cross multiply

296 × 85V2 = 353.5 × 283

25,160V2 = 100,040.5

V2 = 100,040.5 ÷ 25,160

V2 = 3.98dm³

Answer:

Carbon:  14

Oxygen: 16

Nitrogen: 15

Sulpher: 16

A species containing a negatively charged C atom is called a carbanion, and such a species will act readily as a nucleophile

A carbanion can be characterized as an adversely charged particle in which a carbon iota displays trivalence (suggesting it shapes a sum of three bonds) and holds a conventional negative charge whose size is at any rate - 1. Any individual from a class of natural mixtures where a negative electrical charge is found overwhelmingly on a carbon iota.

A nucleophile is a synthetic animal types that structures bonds by giving an electron pair. All particles and particles with a free sets of electrons or possibly one pi bond can go about as nucleophiles. Since nucleophiles give electrons, they are Lewis bases.

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Answer:

Explanation:

When a flask is dipped in very hot water in a large beaker , the flask expands due to heat gain . As a result , level of water in tube fitted in flask goes down .

After some time , the water inside tube also become hot so it expands  . coefficient of volume expansion of water is more than coefficient of volume expansion of glass . Hence greater expansion takes place in the volume of water . It is due to this fact that water level in tube starts rising after some time of fall .

The name of all functional groups in the molecule is carbonyl, alkyne group, thiol, and an amine group.

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The new product that Margerie made was a mixture because of option B. Yes, because she combined two liquids and made a new product that is evenly mixed.

A mixture is a material made of two or extra one-of-a-kind chemical substances which aren't chemically bonded. A combination is a bodily aggregate of extra substances in which the identities are retained and blended within the shape of solutions, suspensions, and colloids.

A combination consists of one or more natural materials in varying compositions. There are two sorts of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, even as homogeneous combos appear uniform all through.

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Answer:

In units of measurement of length we use centimeter (cm) to measure. We can use this unit for measuring the length of a pencil, the width of a book etc. but this unit is too big to measure the thicken of a pencil. So we use another unit called millimeter (mm).

Explanation:

The Lmin, tox, μ, and Vt, we have found the oxide charge density and permittivity of SiO2, the value 0.125V.

Given: Lmin = 0.36 μm

Tox = 4 nmμ = 450 cm2

VsVt = 0.5 V

We have to find Vox.

To find Vox, we will use the following formula: Vox = [Qox/εox]  where Qox is the oxide charge density, and εox is the permittivity of SiO2.

For this calculation, we will use the following formula:.

Tox = εox * tox

So, εox = Tox / tox= 4 nm / 10 nm⁻⁹ = 4×10⁹ F/m

Now, we will find the oxide charge density Qox using the following formula: Qox = Cox * Vtwhere Cox is the oxide capacitance per unit area

Cox = εox / toxCox = (4×10⁹ F/m) / (4×10⁻⁹ m)Cox = 1 F/m²Vox = [Qox/εox]= [Cox * Vt/εox]= [(1 F/m²) * 0.5 V] / (4×10⁹ F/m)= 1.25 × 10⁻¹¹ m= 1.25 × 10⁻¹¹ / 1 × 10⁻⁹= 0.125 V

Explanation:

Given the Lmin, tox, μ, and Vt, we have found the oxide charge density and permittivity of SiO2 using the given formulas.

We then applied the formula to find Vox, and we got the value 0.125V.

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The Lmin, tox, μ, and Vt, we have found the oxide charge density and permittivity of SiO2, the value 0.125V.

Given: Lmin = 0.36 μm

Tox = 4 nmμ = 450 cm2

VsVt = 0.5 V

We have to find Vox.

To find Vox, we will use the following formula: Vox = [Qox/εox]  where Qox is the oxide charge density, and εox is the permittivity of SiO2.

For this calculation, we will use the following formula:.

Tox = εox * tox

So, εox = Tox / tox= 4 nm / 10 nm⁻⁹ = 4×10⁹ F/m

Now, we will find the oxide charge density Qox using the following formula: Qox = Cox * Vtwhere Cox is the oxide capacitance per unit area

Cox = εox / toxCox = (4×10⁹ F/m) / (4×10⁻⁹ m)Cox = 1 F/m²Vox = [Qox/εox]= [Cox * Vt/εox]= [(1 F/m²) * 0.5 V] / (4×10⁹ F/m)= 1.25 × 10⁻¹¹ m= 1.25 × 10⁻¹¹ / 1 × 10⁻⁹= 0.125 V

Explanation:

Given the Lmin, tox, μ, and Vt, we have found the oxide charge density and permittivity of SiO2 using the given formulas.

We then applied the formula to find Vox, and we got the value 0.125V.

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The entropy change for the surroundings when 2.36 moles of \(CaCO_3\)react at standard conditions is -1419.9 J/K.

The standard thermodynamic data at 298 K for the given reaction are:

ΔH° = +179.1 kJ/mol (enthalpy change)

ΔS° = +160.5 J/(mol·K) (entropy change)

To calculate the entropy change for the surroundings, we need to first calculate the amount of heat released (or absorbed) by the reaction. This can be done using the enthalpy change and the number of moles of \(CaCO_3\) reacted:

ΔH = nΔH° = (2.36 mol)(+179.1 kJ/mol) = +422.276 kJ

Since the reaction is exothermic (ΔH is negative), heat is released to the surroundings, and the entropy change for the surroundings is given by:

ΔS = -ΔH/T

where T is the temperature in Kelvin. At standard conditions, T = 298 K, so:

ΔS = -(+422.276 kJ)/(298 K) = -1419.9 J/K

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Answer:

New volume V2 = 120.4 ml

Explanation:

Given:

Pressure P1 = 1.12 atm

New pressure P2 = 1.12[1+0.25] = 1.4 atm

Old volume V1 = 150.5 ml

Find:

New volume V2

Computation:

P1V1 = P2V2

(1.12)(150.5) = (1.4)(V2)

V2 = 120.4 ml

New volume V2 = 120.4 ml

The end of a very long 5-mm-diameter rod is held at 124°C. The surface of the rod is exposed to ambient air at 30°C, with a convective heat transfer coefficient of 100 W/m2K. Determine the temperature at a radial distance of 2.5 mm from the rod's center. The thermal conductivity of the rod is 15 W/mK.b) What is the temperature gradient in the rod at this location?c).

What is the heat flux at this location?The temperature at a radial distance of 2.5 mm from the rod's center is 79.58°C.The solution for this problem can be found by following the steps below:Solution:a) The temperature of the rod, T, can be calculated using the formula for one-dimensional conduction:q/A = -k (d T/d r)whereq is the heat flux,A is the cross-sectional area of the rod,r is the radial distance from the center of the rod,k is the thermal conductivity of the rod,and T is the temperature of the rod.

Taking the boundary condition into account,T(r=0) = 124°CandT(r=2.5 mm) = 30°C, the solution to the differential equation is:T = T0 + (T1 - T0) (r/R)2whereT0 = 30°CT1 = 124°CR = 2.5 mm/2 = 1.25 mmso,T = 30 + (124 - 30) (r/1.25)2 = 30 + 78 (r/1.25)2at r = 2.5 mm,T = 79.58°Cb) The temperature gradient, d T/d r, is given by the derivative of the above equation:d T/d r = 124 (r/1.25)2 / 1.25where d T/d r = 98.72°C/mat r = 2.5 mmc) The heat flux, q/A, is given by the Fourier's law of heat conduction:q/A = -k (d T/d r)whereq/A = -15 (98.72/1000) = -1.48 W/m2at r = 2.5 mmTherefore, the temperature at a radial distance of 2.5 mm from the rod's center is 79.58°C, the temperature gradient in the rod at this location is 98.72°C/m, and the heat flux at this location is -1.48 W/m2.

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The equilibrium constant (K) for the given reaction at 1000 K and 1 bar is X. The maximum possible conversion for an equimolar feed is Y.

The equilibrium constant (K) is a measure of the extent to which a reaction reaches equilibrium. It is defined as the ratio of the concentrations (or partial pressures) of the products to the concentrations (or partial pressures) of the reactants, each raised to the power of their respective stoichiometric coefficients. In this case, the equilibrium constant (K) can be determined by considering the balanced chemical equation:

C₂H4(g) + H₂O(g) → C₂H5OH(g)

The equilibrium constant expression is given by: K = [C₂H5OH] / [C₂H4] [H₂O]

To calculate the maximum conversion for an equimolar feed, we need to consider the stoichiometry of the reaction. Since the reactants are in equimolar proportions, we can assume that the initial concentration of C₂H4 is equal to the initial concentration of H₂O.

Now, let's calculate the equilibrium constant and maximum conversion based on the provided information.

equilibrium constant and maximum conversion:

The equilibrium constant (K) provides information about the position of a chemical reaction at equilibrium. It indicates the relative concentrations of products and reactants when the reaction reaches a state of balance. A high value of K suggests that the reaction favors the formation of products, while a low value indicates a preference for the reactants.

In this particular case, we are given the reaction C₂H4(g) + H₂O(g) → C₂H5OH(g) at 1000 K and 1 bar. To calculate the equilibrium constant (K), we compare the concentrations (or partial pressures) of the products (C₂H5OH) and reactants (C₂H4 and H₂O). The equilibrium constant is a dimensionless quantity that quantifies the equilibrium position.

To determine the maximum conversion for an equimolar feed, we consider the stoichiometry of the reaction. Since the reactants are in equimolar proportions, it means that the initial concentration of C₂H4 is equal to the initial concentration of H₂O. The maximum conversion refers to the maximum extent to which the reactants can be converted into products under the given conditions.

By solving the equilibrium constant expression and considering the stoichiometry, we can calculate both the equilibrium constant and the maximum conversion for this reaction.

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The energy of a photon : 3 x 10⁺²⁰

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

\(\large{\boxed{\bold{E\:=\:h\:.\:f}}}\)

Where

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves

f = c / λ

c = speed of light

= 3.10⁸

λ = wavelength

Wavelength of photon=λ=6 μm = 6.10⁻⁶ m

\(\tt E=h\times \dfrac{c}{\lambda}\\\\E=6,626.10^{-34}\times \dfrac{3.10^8}{6.10^{-6}}\\\\E=3.3\times 10^{-20}~J\)

The characteristics and the type of molecular bond they describe:

1. Bonds found in Halite (between Na⁺ and Cl⁻): Ionic bond

2. Bonds found between Si and O in the Si-O tetrahedron: Covalent bond

3. Bonds inside the water molecule (between the H and O): Covalent bond

4. Bonds that exist between two water molecules: Hydrogen bond

5. Strongest bond type: Covalent bond

6. Weakest bond type: Van der Waals bond

7. Bonds that are used by water to dissolve salt: Ionic bond

The ionic bond is a type of molecular bond found in halite (between Na⁺ and Cl⁻). The Si-O tetrahedron is held together by a covalent bond. The bond inside the water molecule (between the H and O) is also a covalent bond. The hydrogen bond is the type of bond that exists between two water molecules. The covalent bond is the strongest bond type, while the van der Waals bond is the weakest bond type. Water uses the ionic bond to dissolve the salt.

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In any aqueous solution at 25°C, the ionic-product of water (Kw) is a constant value equal to 1.0 x 10⁻¹⁴ mol²/L².

Due to the auto-ionization or self-ionization of water, water molecules dissociate into hydronium ions (H₃O⁺) and hydroxide ions (OH⁻). At 25°C, the Kw is equal to 1.0 x 10⁻¹⁴ mol²/L².

This constant value of Kw plays a crucial role in understanding the acidity and basicity of aqueous solutions. It helps to establish the relationship between the concentrations of hydronium and hydroxide ions, as their product remains constant at a given temperature. The pH and pOH scales are derived from this relationship, providing a convenient method for measuring the acidity or basicity of a solution.

In summary, the ionic-product of water, Kw, remains constant at 1.0 x 10⁻¹⁴ mol²/L² for any aqueous solution at 25°C. This constant is a result of the auto-ionization of water and helps to understand the relationship between hydronium and hydroxide ions in the context of acidity and basicity.

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