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A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answers

Answer 1

Answer:

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\nV_1=160.0mL\nM_2=0.0501M\nV_2=?$

Putting values in above equation, we get:

$0.3403M* 160.0mL=0.0501M* V_2\n\nV_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=(1)/(2)[pK_w+pK_b+\log C]$

$pOH=(1)/(2)[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\n\npH=14-pOH\n\npH=14-8.76\n\npH=5.24$

Thus, the pH of the solution is, 5.24

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Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Answers

The number of oxygen atoms in 19.3 g of sodium sulfate (Na₂SO₄) is 3.27×10²³ atoms

We'll begin by calculating the number of mole in 19.3 g of sodium sulfate (Na₂SO₄).

Mass of Na₂SO₄ = 19.3 g

Molar mass of Na₂SO₄ = (23×2) + 32 +(16×4)

= 46 + 32 + 64

= 142 g/mol

Mole of Na₂SO₄ =?

Mole = mass / molar mass

Mole of Na₂SO₄ = 19.3 / 142

Mole of Na₂SO₄ = 0.136 mole

Recall:

1 mole of Na₂SO₄ contains 4 moles of O.

Therefore,

0.136 mole of Na₂SO₄ will contain = 0.136 × 4 = 0.544 mole of O

Finally, we shall determine the number of atoms in 0.544 mole of O.

From Avogadro's hypothesis,

1 mole of O = 6.02×10²³ atoms

Therefore,

0.544 mole of O = 0.544 × 6.02×10²³

0.544 mole of O = 3.27×10²³ atoms

Thus, 19.3 g of sodium sulfate (Na₂SO₄) contains 3.27×10²³ atoms of oxygen.

Learn more: brainly.com/question/25115547

Answer:

3.27·10²³ atoms of O

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is Na₂SO₄, and its molar mass is approximately 142.05 $(g)/(mol)$ .

We will use stoichiometry to convert from our mass of Na₂SO₄ to moles of Na₂SO₄, and then from moles of Na₂SO₄ to moles of O using the mole ratio; then finally, we will convert from moles of O to atoms of O using Avogadro's constant.

19.3g Na₂SO₄ · $(1 mol Na^2SO^4)/(142.05g Na^2SO^4)$ · $(4 mol O)/(1 mol Na^2SO^4)$ · $(6.022x10^2^3)/(1 mol O)$

After doing the math for this dimensional analysis, you should get a quantity of approximately 3.27·10²³ atoms of O.

what is the frequency of light emitted when the electron in a hydrogen atom undergoes a transition from energy level n

Answers

Complete question is;

What is the frequency of light emitted when the electron in a hydrogen atom undergoes a transition from energy level n=6 to level n=3?

Answer:

Frequency = 2.742 × 10^(14) s^(-1)

Explanation:

First of all, the energy of hydrogen electron from online values is;

E_n = -2.18 × 10^(-18) × (1/n²) J

n is the principal quantum number

We are told that hydrogen atom undergoes a transition from energy levels n = 3 to n = 6.

Thus, it means we have to find the difference between the electrons energy in the energy levels n = 3 to n = 6.

Thus;

E_n = E_6 - E_3

Thus;

E_n = [-2.18 × 10^(-18) × (1/6²)] - [-2.18 × 10^(-18) × (1/3²)]

E_n = (2.18 × 10^(-18)) × [-1/36 + 1/9]

E_n = 0.1817 × 10^(-18) J

From Planck expression, we can find the frequency. Thus;

E = hf

Where h is Planck's constant = 6.626 × 10^(-34) m²kg/s

Thus;

0.1817 × 10^(-18) = 6.626 × 10^(-34) × f

f = (0.1817 × 10^(-18))/(6.626 × 10^(-34))

f = 2.742 × 10^(14) s^(-1)

Final answer:

The frequency of light emitted during an electron transition in a hydrogen atom is determined by calculating the energy difference between the two energy levels and then using this to calculate the frequency using the equation for energy of a photon.

Explanation:

The frequency of light emitted during a transition of an electron in a hydrogen atom can be calculated using the formula for the energy difference (∆E) between two energy levels n1 and n2 in the hydrogen energy level diagram.

The formula to calculate energy difference is: ∆E = E(n2) - E(n1) where E(n) represents the energy of an energy level n. The energy difference ∆E is negative when an electron goes down an energy level (i.e., emits a photon), as the energy level n1 is greater than n2.

The frequency of the emitted photon (∆E) is then given by the formula ∆E = hf where h is Planck's constant (6.63 x 10^-34 Joule seconds) and f is the frequency. Therefore, you can rearrange the equation to find the frequency: f = ∆E / h.

Learn more about Electron Transitions here:

brainly.com/question/8384785

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Does the molarity of the solution change with the change in temperature?Match the words in the left column to the appropriate blanks in the sentences on the right.

Answers

Answer:

Yes, molarity changes according to temperature.

Explanation:

Molarity equals to moles over volume

which can be also said;

Molarity = moles of solute / liters of solution

Well, the volume or liters of the solution depends on temperature.

If the temperature is high then the volume will be higher where the molarity decreases.

One way to experimentally measure the heat capacity of a Styrofoam cup calorimeter would be to melt a known mass of ice in warm water and measure hte temperature change. Use the data below to determine the experimental heat capacity of the calorimeter. Use the literature heat of fusion for ice in your calculations. Assume the ice added is at 0.00 c.Mass of ice added: 17.69gMass of water in calorimeter: 98.67gT-Initial of water: 28.7T-Final of water after melting ice: 12.9C

Answers

Answer:

4.88 Cals per degree celsius

Explanation:

We have taken heat of fusion of ice = 80 cals / g

We have taken speciic heat of water = 1 cal/g per degree celsius

In this experiment , let the heat capacity of calorimeter be X.

Heat gained by ice

heat gained in melting + heat gained in getting warmed

= mass x latent heat + mass x specific heat x rise in temperature

= 17.69 x 80 + 17.69 x 1 x ( 12.9 - 0 )

= 1643.4 Cals

Heat lost by water

= mass x specific heat x fall in temperature

98.67 x 1 x ( 28.77 - 12.9 )

= 1565.89 Cals

Heat lost by calorimeter

heat capacity x fall in temperature

X x ( 28.77 - 12.9 )

Heat gained = heat lost

1643.4 = 1565.89 +15.87X

X = 4.88 Cals per degree celsius

What bond distance is expected to be longest?1. A carbon-carbon bond with a bond order of 2
2. A carbon-carbon bond with a bond order of 3
3. A carbon-carbon bond with a bond order of O
4. carbon-carbon bond with a bond order of 1

Answers

Answer:

Bond length of C=C is largest(134 pm) because both the carbon atoms have same electronegativity. In case of C=O. and C=N carbon is bonded to highly electronegative atoms so bond length is shoreter as compared to C=C

Which of the following is an example of a compound? water - H2Ooxygen-O2
hydrogen - H2
helium - He

Answers

Answer:

water

Explanation:

the other options are elements while water is 2 elements

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