A 75.0 ml volume of 0.200 m nh3 (kb = 1.8 * 10^-5) is titration with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3

Answer:

top one is anaphase and the middle one is cytokinesis can't see enough of the bottom one to tell

Explanation:

a) The temperature of the air after passing through the heating section is 28°C, and the relative humidity is 50%.

b) The rate of heat transferred to the air is 452.25 kW.

c) The rate of water added by the evaporative cooler is 0.305 kg/min or 18.3 kg/h.

d) The evaporative cooler is important because it adds moisture to the air.

To determine the temperature and relative humidity of the air after it passes through the heating section, we can follow the following steps:

a) Temperature and Relative Humidity Calculation:

The air enters the evaporative cooler at 10°C with a relative humidity of 60%.

We need to calculate the water vapor pressure (Pw) using the saturation vapor pressure at 10°C, which is 1.227 kPa.

Pw = 60% x 1.227 = 0.736 kPa

The remaining partial pressure of the air (Pa) can be calculated by subtracting the water vapor pressure from the atmospheric pressure (101.3 kPa).

Pa = 101.3 - 0.736 = 100.56 kPa

Next, the air enters the heating section at 28°C. We need to calculate the water vapor pressure using the saturation vapor pressure at 28°C, which is 3.733 kPa.

Pw = 50% x 3.733 = 1.866 kPa

Again, calculate the remaining partial pressure of the air (Pa) by subtracting the water vapor pressure from the atmospheric pressure (101.3 kPa).

Pa = 101.3 - 1.866 = 99.43 kPa

Therefore, the temperature of the air after passing through the heating section is 28°C, and the relative humidity is 50%.

b) Rate of Heat Transfer Calculation:

The mass flow rate of the air is given as 25 m³/min.

We can convert the mass flow rate to kg/min by using the density of air.

Mass flow rate = Volume flow rate x Density

Density of air is approximately 1.225 kg/m³.

Mass flow rate = 25 x 1.225 = 30.625 kg/min

The specific heat capacity of air at constant pressure is 1.005 kJ/(kg·K).

Now we can calculate the rate of heat transferred using the formula:

Rate of heat transferred = mass flow rate x specific heat capacity x (Tout - Tin)

Tin = 10°C (temperature of air entering the evaporative cooler)

Tout = 28°C (temperature of air leaving the evaporative cooler)

Rate of heat transferred = 30.625 x 1.005 x (28 - 10) = 452.25 kW

Therefore, the rate of heat transferred to the air is 452.25 kW.

c) Rate of Water Added Calculation:

The mass flow rate of the dry air is the same as the mass flow rate of the air, which is 25 kg/min.

To calculate the specific humidity at the inlet and outlet, we need to determine the mass of water vapor and the mass of dry air.

Specific humidity at inlet = Pw / (Pa - Pw)

Pw = 0.736 kPa (calculated earlier)

Pa = 100.56 kPa (calculated earlier)

Specific humidity at outlet = Pw / (Pa - Pw)

Pw = 1.866 kPa (calculated earlier)

Pa = 99.43 kPa (calculated earlier)

Now we can calculate the rate of water added using the formula:

Rate of water added = mass flow rate of dry air x (specific humidity outlet - specific humidity inlet)

Rate of water added = 25 x (0.0193 - 0.0074) = 0.305 kg/min or 18.3 kg/h

Therefore, the rate of water added by the evaporative cooler is 0.305 kg/min or 18.3 kg/h.

d) Importance of Evaporative Cooler:

The evaporative cooler plays an important role in the air conditioning system because it adds moisture to the air. When hot and dry air enters the evaporative cooler, it passes through water-saturated pads that absorb moisture from the water and add it to the air. This process cools and humidifies the air, making it more comfortable to breathe. By increasing the relative humidity, the evaporative cooler helps to alleviate dryness and provides a more pleasant environment.

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Answer:

I believe the answer is a. I hope this helps

The type of electrons that are best at shielding a 3p electron is 4) 3s electrons.

Shielding refers to the ability of electrons in inner energy levels to repel or shield outer electrons from the full effect of the positive charge of the nucleus. Electrons in lower energy levels (closer to the nucleus) have a stronger shielding effect on outer electrons.

In this case, the 3p electron is in the outermost energy level. The electrons in the 3s sublevel are in the same energy level as the 3p electron but are closer to the nucleus. Therefore, the 3s electrons have a better shielding effect on the 3p electron compared to the other options listed.

The 2p electrons (option 1) are in a lower energy level, so they have less shielding effect on the 3p electron. The 3p electrons themselves (option 2) do not contribute to the shielding effect. The 4p electrons (option 3) are in a higher energy level and are further away from the nucleus, so their shielding effect is weaker. The 3d electrons (option 5) are in a higher energy level but have less shielding effect compared to the 3s electrons.

Therefore, the 3s electrons (option 4) are best at shielding a 3p electron.

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Answer:

N2+ 3H2－> 2NH3

2Fe + 6HCl －> 3H2 + 2FeCl3

The model used to determine molecular shape based on minimizing the repulsion of shared and unshared electron pairs around the central atom is the Valence Shell Electron Pair Repulsion (VSEPR) theory.

According to the VSEPR theory, electron pairs (both bonding and nonbonding) around the central atom repel each other and tend to position themselves as far apart as possible to minimize this repulsion. By considering the number of bonding and nonbonding electron pairs, the VSEPR theory predicts the molecular geometry or shape of a molecule.

The VSEPR theory allows us to determine whether the molecular shape is linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, or other shapes, depending on the arrangement of electron pairs. This model is widely used to understand and predict the shapes of molecules and their properties.

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The refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.

To calculate the kilowatt-hours (kWh) of electric energy used by the refrigerator in 30 days, we need to multiply the power rating by the total running time.

Given:

Power rating of the refrigerator = 500 W

Running time per day = 12 hours

Number of days = 30

First, we need to convert the power rating from watts to kilowatts:

Power rating = 500 W / 1000 = 0.5 kW

Next, we calculate the total energy used in kilowatt-hours (kWh) over the 30-day period:

Energy used = Power rating × Running time × Number of days

Energy used = 0.5 kW × 12 hours/day × 30 days

Energy used = 180 kWh

Therefore, the refrigerator would use 180 kilowatt-hours (kWh) of electric energy over the course of 30 days, assuming it runs for 12 hours each day.

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Answer:

47.6°C is the change of temperature

Explanation:

To solve this question we must use the equation of specific heat of a material:

Q = m*ΔT*S

Where Q is heat applied = 90J

m is the mass of the substance = 15g gold

ΔT is change in temperature = Our incognite

S is specific heat of the material = 0.126J/g°C for gold

Replacing:

90J = 15g*ΔT*0.126J/g°C

90J/15g*0.126J/g°C = ΔT

ΔT = 47.6°C is the change of temperature

The melting and boiling and density is given below :

Substance          melting point / °C         boiling point / °C       density / g cm⁻³

  E                           -259                                 -253                          0.09

 F                           1085                                 2580                         8.93

  G                                 -7                                     59                            3.1

 H                              -39                                   357                           13.6

  I                                -218                                  - 183                       0.0013

A) substances are gases at room temperature:

    E and I are gases at room temperature because the melting and boiling point are below at room temperature.

B) substances are liquids at room temperature:

    G and H  are liquids at room temperature because substances are liquid at room temperature if melting point is below room temperature and boiling point is above room temperature.

C) substances are solid at room temperature :

   F is solid at room temperature because its melting point is above room temperature.

D) substances are most likely to be metal :

   F is metal because metals have very high melting and boiling point except mercury.

E) substance are most likely to be mercury :

    H is most likely to be mercury because mercury has boiling point -38°C and melting point is 356°C and density is 13.5 g/ cm³ and mercury is metal with exception.

G) substances will be liquid at -210°C :

  I is liquid at -210°C because substance is liquid when its melting point is below temperature and boiling point is above temperature.

H) substances are most likely to be bromine :

  G is more likely to be bromine because melting and boiling point and density of G is close to bromine.

I)  substances which are least dense non metal :

   E is least dense non metal because its melting and boiling point and density is close to least dense non - metal.

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The pKa of acetic acid is 4.74, 6 M acetic acid to prepare 100 mL of 0.10 M \(CH3COOH\) and 9.09 mL of 0.10 M \(CH_2COOH\) and 0.91 mL of 0.10 M CH3COONa.

1. The pKa of acetic acid is 4.74, which means its buffering range is between pH 4.74 and pH 5.74. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Acetic acid and its conjugate base, acetate, combine in solution to form a buffer. The amount of acetic acid and acetate will remain relatively constant over this pH range.

2. To prepare a 0.10 M \(CH3COOH\) solution, you need to calculate the volume of 6 M acetic acid needed.

Using the formula: \(M1V1 = M2V2\), you can calculate that you need 0.17 mL of 6 M acetic acid to prepare 100 mL of 0.10 M \(CH3COOH\).

3. To prepare 10.0 mL of buffer at the following pH values, you need to calculate the volume of 0.10 M\(CH_2COOH\) and 0.10 M \(CH3COONa\). At pH 3.7, you will need 4.74 mL of 0.10 M CH2COOH and 5.26 mL of 0.10 M CH3COONa. At pH 4.7, you will need 6.45 mL of 0.10 M CH2COOH and 3.55 mL of 0.10 M CH3COONa. At pH 5.7, you will need 9.09 mL of 0.10 M CH2COOH and 0.91 mL of 0.10 M CH3COONa.

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1. The pKa of acetic acid is 4.74, which means its buffering range is between pH 4.74 and pH 5.74. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Acetic acid and its conjugate base, acetate, combine in solution to form a buffer. The amount of acetic acid and acetate will remain relatively constant over this pH range.

2. To prepare a 0.10 M CH3COOH solution, you need to calculate the volume of 6 M acetic acid needed. Using the formula: M1V1 = M2V2, you can calculate that you need 0.17 mL of 6 M acetic acid to prepare 100 mL of 0.10 M CH3COOH.

3. To prepare 10.0 mL of buffer at the following pH values, you need to calculate the volume of 0.10 M CH2COOH and 0.10 M CH3COONa. At pH 3.7, you will need 4.74 mL of 0.10 M CH2COOH and 5.26 mL of 0.10 M CH3COONa. At pH 4.7, you will need 6.45 mL of 0.10 M CH2COOH and 3.55 mL of 0.10 M CH3COONa. At pH 5.7, you will need 9.09 mL of 0.10 M CH2COOH and 0.91 mL of 0.10 M CH3COONa.

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Answer:

1) 60°

2) 60°

3) 45°

4) E

5)90°

6) E

Given data:

Mole% of benzene (B) = 30%

Mole% of toluene (T) = 25%

Balance = xylene (X)

Bottom product contains mole% of X = 98%

The overhead product from the second column contains mole% of B in the feed to this column = 97%Composition of overhead product = 94% B

Mole% of X recovered in the bottoms product = 96%

Let's solve the problem:Since the sum of mole% of benzene, toluene, and xylene is 100%, the mole% of X = 100 - 30 - 25 = 45%

In the feed, mole% of B = 30%, mole% of T = 25%, and mole% of X = 45%.

Calculation of benzene emerging in the overhead product from the second column:

Let's consider the flow rate of feed as 100 units, so the flow rate of B in the feed = 100 × 0.3 = 30 units.The flow rate of B in the overhead product of the first column = 30 × 0.03 = 0.9 units (since 97% of benzene in the feed is recovered in the overhead product).

Thus, the flow rate of benzene in the feed to the second column is 0.9 units.The composition of the overhead product from the second column is 94% benzene. Thus, the flow rate of benzene in the overhead product from the second column = 0.9 × 0.94 = 0.846 units.

The percentage of benzene emerging in the overhead product from the second column = (0.846/30) × 100 = 2.82%.

Hence, the percentage of benzene in the process feed that emerges in the overhead product from the second column is 2.82%.Calculation of toluene emerging in the bottom product from the second column:

Let's consider the flow rate of feed as 100 units, so the flow rate of T in the feed = 100 × 0.25 = 25 units.The flow rate of T in the overhead product from the first column = 25 × 0.04 = 1 unit (since 96% of the X in the feed is recovered in the bottoms product).

Thus, the flow rate of T in the feed to the second column is 25 - 1 = 24 units.The composition of the bottom product from the second column is 98% X. Thus, the flow rate of X in the bottom product from the second column = 24 × 0.98 = 23.52 units.

The flow rate of T in the bottom product from the second column = 100 - 30 - 23.52 = 46.48 units.The percentage of toluene in the process feed that emerges in the bottom product from the second column = (46.48/25) × 100 = 185.92%.

Hence, the percentage of toluene in the process feed that emerges in the bottom product from the second column is 185.92%.

Thus, the required answers are:

(i) The percentage of benzene in the process feed that emerges in the overhead product from the second column = 2.82%

(ii) The percentage of toluene in the process feed that emerges in the bottom product from the second column = 185.92%.

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Answer:

2KNO3 + H2SO4 → K2SO4 + 2HNO3

Explanation:

Hope this helps

Answer:

everything can be found in the picture

There are 293 milligrams of the solute in 315 mL of the given solution.

To calculate the mass of solute in milligrams, we need to use the following formula:

mass of solute (in mg) = volume of solution (in mL) x concentration of solute (in ppm) x molecular weight of solute / 10^6

Here's how we can apply this formula to solve the problem:

Convert the given concentration from ppm to g/mL:

2.73 ppm = 2.73 mg/L (since 1 ppm = 1 mg/L)

= 2.73 x 10^-3 g/mL (since 1 mg = 10^-3 g)

Calculate the mass of solute in grams:

mass of solute = 315 mL x 2.73 x 10^-3 g/mL x 331.20 g/mol / 10^6

= 0.293 g

Convert the mass of solute from grams to milligrams:

mass of solute = 0.293 g x 10^3 mg/g

= 293 mg

Therefore, there are 293 milligrams of the solute in 315 mL of the given solution.

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Answer:

the mixture two is the Heterogeneous solution and the mixture 1 (sugar solution) is true solution

Explanation:

Heterogeneous solutions are solutions with non-uniform composition and properties throughout the solution. A solution of oil and water, water and chalk powder and solution of water and sand, etc.

True solution

The sugar solution is a mixture of sugar and water which is a true solution. True Solution is a homogeneous mixture of two or more materials with a particle size of less than 10-9 m or 1 nm dissolved in the solvent.

A solute is a substance to be dissolved.

The solvent is the one in which solute is dissolving.

In the sugar solution sugar is the solute and water is the solvent.

Answer:

Explanation:

your cool

The system at 25°C, the equilibrium will shift to the right. NH4Cl(s) will break down into NH4(aq) and Cl-(aq) forming more products as the temperature increases.

The reaction NH4Cl(s) heat ↔ NH4(aq) + Cl-(aq) is an endothermic reaction where heat is absorbed as the reactant NH4Cl(s) breaks down into NH4(aq) and Cl-(aq) at a specific temperature of 25°C.

The equilibrium will shift to the right if an increase in temperature is added to the system.

The increase in temperature will favour the endothermic reaction that requires more heat for the reaction to occur as the system shifts in the direction that will absorb more heat to replace that lost to the surroundings.

So, when an increase in temperature is added to the system at 25°C, the equilibrium will shift to the right.

Hence, NH4Cl(s) will break down into NH4(aq) and Cl-(aq) forming more products as the temperature increases.

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Answer:

magnesium in biological system,magnesium is smaller atom because calcium has more electron (20e-) which will occupy more energy levels (n=4) for calcium versus n=3 for magnesium)

Explanation:

maybe Brilliant answer?

When HBr is added to an alkene, the major product obtained is the alkyl halide.

The specific product formed depends on the nature of the alkene and the conditions of the reaction. The reaction proceeds through electrophilic addition, where the carbocation acts as an electrophile, and the HBr molecule acts as a nucleophile.

The addition of HBr to an alkene happens through the Markovnikov addition. The nucleophilic \(Br^{-}\) ion adds to the carbon atom bearing the most hydrogen atoms, leading to the formation of an alkyl halide with the halogen (Br) attached to the more substituted carbon. This is known as the Markovnikov addition. This reaction occurs with more stable carbocations.

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Answer:

58,5

Explanation:

Explanation:

more saturated than solution one

Answer:

Electrons are the orbiting particles

Explanation:

Hope this helps UvU

To collect data for the Synthesis of Aspirin experiment, you will measure the mass of salicylic acid, volume of acetic anhydride, mass of aspirin obtained, and the beginning and end of the melting point range for the product.

- Mass of salicylic acid used: 0.152 g
- Volume of acetic anhydride used: 0.36 mL
- Mass of aspirin obtained: 0.176 g
- Beginning of melting point range for product: 125.8 °C
- End of melting point range for product: 131.3 °C
Salicylic acid has a molar mass of 138.12 g/mol, and acetic anhydride has a molar mass of 102.09 g/mol. You can then calculate moles using the formula: moles = mass (g) / molar mass (g/mol).

Hence,  For the Synthesis of Aspirin experiment, collect data by measuring the mass of salicylic acid, volume of acetic anhydride, mass of aspirin obtained, and the beginning and end of the melting point range for the product.

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The value of Ksp for AglO₃, if it is found that up to 0.0980 g of Aglo₃ dissolves in 2.00 L of aqueous solution at a certain temperature, is 0.2625 mol/L

To determine the value of Ksp for AglO₃, we must find the concentration of Ag⁺ ion and IO₃⁻ ion in the saturated solution of AglO₃ be x M. Since one mole of AglO₃ produces one mole of Ag⁺ ion and one mole of IO₃⁻ ion. The balanced chemical equation for the dissociation of AglO₃ is as follows:

AglO₃(s) ⇌ Ag+ (aq) + IO₃⁻(aq)

Initially, the concentration of AglO₃ is x M. As the AglO₃ is completely dissociated in the aqueous solution, the concentration of Ag⁺ ion and IO₃⁻ ion also become x M at the equilibrium. Hence, the equilibrium concentration of Ag⁺ ion and IO₃⁻ ion is x M.

The Ksp expression for the dissociation of AglO₃ is as follows:

Ksp = [Ag⁺][IO₃⁻]

Ksp = x × x = x²

Ksp = 0.0980 g/ (2.00 L) × 1000 g/ 1 kg × 1 mol/ 187.81 g

= 0.2625 mol/L (approx)

Thus, the value of Ksp for AglO₃ is 0.2625 mol/L.

ICE table is shown below:

              AglO₃(s) ⇌ Ag⁺ (aq) + IO₃⁻(aq)

Initial (M)       x              0                 0

Change (M) -x              +x               +x

Equilibrium (M) x          x                 x

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Based on the empirical formula of C2H3O, the molecular formula of the compound can be calculated using its molecular mass:

Molecular formula mass = (empirical formula mass) x n

where n is a positive integer that represents the number of empirical formula units in the molecule.

For this compound, the molecular formula mass is 86 amu, which is the same as the molecular mass given in the problem. Therefore, n = 1, and the molecular formula is also C2H3O.

Plausible structure with alcohol, ether, and alkyne functional groups:

One possible structure for this compound that contains an alcohol, ether, and alkyne functional groups is propargyl alcohol.

H C≡C─CH2─OH

│

O

This compound has the empirical formula C2H3O and a molecular mass of 56 amu. To get the molecular mass of 86 amu, we can add a methyl group (CH3) to the propargyl alcohol molecule:

H C≡C─CH2─O─CH3

│

O

Plausible structure with aldehyde and ketone functional groups:

Another possible structure for this compound that contains an aldehyde and ketone functional groups is propanal acetone.

H

│

O═C─CH2─CH3

│

O

│

CH3

This compound has the empirical formula C2H3O and a molecular mass of 86 amu.

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The initial charge on each sphere is: q1 = 0.0438 N q2 = 1.283 * 0.0438 N = 0.0562 N

To find the initial charge on each sphere, we can use the equation for the electrostatic force between two charged spheres:

F = (k * |q1 * q2|) / \(r^2\)

where F is the force, k is the Coulomb force constant, q1 and q2 are the charges on the spheres, and r is the distance between the spheres.

Given that F1 = 0.0780 N and F2 = 0.100 N, and the spheres are identical, we can set up the following equations:

\(0.0780 = (k * |q1 * q2|) / (0.302)^2 ...(1)\\0.100 = (k * |q1 * q2|) / (0.302)^2 ...(2)\)

Dividing equation (2) by equation (1), we get:

0.100 / 0.0780 = (k * |q1 * q2|) / (k * |q1 * q2|)

0.100 / 0.0780 = 1

This tells us that F2 is 1.282 times F1.

Since the spheres are identical, we can assume that the ratio of the charges on the spheres is the square root of the ratio of the forces:

sqrt(q2/q1) = sqrt(F2/F1) = sqrt(1.282) = 1.133

Squaring both sides of the equation, we get:

q2/q1 = \((1.133)^2\) = 1.283

Since q1 is initially less than q2, we can assign a value of q1 to be x, and q2 to be 1.283x.

Now we can solve for the values of q1 and q2:

q1 + q2 = x + 1.283x = 2.283x = 0.100 N (from F2)

Solving for x, we find:

x = 0.100 N / 2.283 = 0.0438 N

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