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wlad13
24 days ago
13

To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water

is mixed with 5 mL of phenol solution and 2 mL of sodium hypochlorite solution, and diluted to 25.0 mL in a volumetric flask. In sample B, 10.0 mL of lake water is mixed with 5 mL of phenol solution, 2 mL of sodium hypochlorite solution, and 2.50 mL of a 5.50×10−4 M ammonia solution, and diluted to 25.0 mL. Sample C is a reagent blank. It contains 10.0 mL of distilled water, 5 mL of phenol solution, and 2 mL of sodium hypochlorite solution, diluted to 25.0 mL. The absorbance of the three samples is then measured at 625 nm in a 1.00 cm cuvet. The results are shown in the table.Sample Absorbance (625 nm)A 0.419B 0.666C 0.045What is the molar absorptivity (????) of the indophenol product?????=M−1cm−1What is the concentration of ammonia in the lake water?[NH3]lake water=M
Chemistry
1 answer:
eduard [2.5K]24 days ago
4 0

Answer:Sample Absorbance (625 nm)

A 0.536

B 0.783

C 0.045

Thus, I will make use of this information to address your inquiry. If you possess different absorbance values, simply follow my approach and substitute other figures.

Initially, we note that 1 mole of NH3 yields 1 mole of product.

For moles of NH3 in B = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) volume of B = 25 = volume of A

Now A = el C = eC (as l = 1cm)

Because the net absorbance due to the complex must account for blank absorbance.

Thus, A(A) = 0.536 - 0.045 = 0.491, A(B) = 0.783 - 0.045 = 0.738

(you can substitute different figures in this stage)

A2/A1 = C2/C1, A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

Thus, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Consequently, [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Volume of lake water = 10 ml out of 25,

Ammonia concentration in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

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Calculate the mass, in grams, of a single tellurium atom (mte = 127.60 amu ).
castortr0y [2731]

1 atomic mass unit (amu) represents the mass of an atom or is used to measure mass on an atomic scale. It is also referred to as a dalton, abbreviated as Da, while atomic mass unit is indicated as amu.

1 amu can be translated into grams as follows:

1 amu = 1.6 * 10^-2^4 g

Mass of Te = 127.6 amu

For conversion into grams:

M = (127.6 ) * 1.6 * 10^-2^4 g

M = 204.16 * 10^-2^4 g

Therefore, the mass of Te is 204.16 * 10^-2^4 g

8 0
1 month ago
Read 2 more answers
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
eduard [2509]

Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = h \nu

1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

\nu = 0.163 \times 10^{17} s^{-1}

or, \nu = 1.63 \times 10^{16} s^{-1}

It is known that \nu = \frac{c}{\lambda}

1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}

\lambda = 1.84 \times 10^{-8} m

According to the De-Broglie equation \lambda = \frac{h}{p}

with p = m \times \nu

So, \lambda = \frac{h}{m \times \nu}

m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m} = 3.6 \times 10^{-26} J/m

Squaring both sides gives us:

(m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}

12.96 \times 10^{-52} = m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where m = mass of the electron

Therefore, m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

=\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

=1.42 \times 10^{-21} J

Since K.E = \frac{1}{2}m \nu^{2}

= \frac{1.42 \times 10^{-21} J}{2}

=0.71 \times 10^{-21} J

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is 7.1 \times 10^{-22} J.

4 0
25 days ago
The density of an alcohol is 0.788 g/mL. What volume in microliters, μL, will correspond to a mass of 20.500 mg?
lions [2649]

Answer:

B.  26.0 μL.

Explanation:

Hello,

Considering the provided mass and density, the volume calculates to be:

V=\frac{m}{\rho} =\frac{25.000mg}{0.788g/mL}*\frac{1g}{1000mg}*\frac{1000\mu L}{1mL} \\ \\V=26.0\mu L

Thus, the solution is B.  26.0 μL.

Best regards.

6 0
13 days ago
A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press
Tems11 [2390]

Answer: The vapor pressure of naphthalene within the flask remains at 2.906\times 10^{-4} atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)

  • The formula employed to determine the enthalpy change for the reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})

The provided information includes:

\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol

Substituting the values into the previous equation produces:

\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol

  • The formula utilized to compute Gibbs free energy change is of a reaction:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]

The equation for the enthalpy change for the reaction is:

\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})

The given factors include:

\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol

By inserting values from the above equation, we arrive at:

\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol

  • For the calculation of K_1 (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

\Delta G^o=-RT\ln K_1

where,

\Delta G^o = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_1 = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}

  • To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})

where,

K_2 = Equilibrium constant at 35°C =?

K_1 = Equilibrium constant at 25°C = 1.14\times 10^{-4}

\Delta H = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = 8.314J/K mol

T_1 = Initial temperature = 25^oC=[273+25]K=298K

T_2 = Final temperature = 35^oC=[273+35]K=308K

By plugging values into the equation above, we obtain:

\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}

  • In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for K_p, which is:

K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for K_2 will equal K_p

p_{C_{10}H_8}=2.906\times 10^{-4}

Consequently, the partial pressure of naphthalene at 35°C is 2.906\times 10^{-4} atm.

3 0
8 days ago
Magnesium metal burns with a bright white flame. What conclusions can you draw about the electron transitions that can take plac
alisha [2704]

Different wavelengths are involved.

Explanation:

When magnesium ignites with a bright white flame, it indicates that various wavelengths are related to the electron transitions occurring in the magnesium atom.

  • Upon combustion, the electrons within the atom become excited.
  • They emit characteristic light that corresponds to their energy levels.
  • White light consists of a mix of different wavelengths.
  • Seeing white light implies that multiple wavelengths combined are responsible for the observed emission.

Learn more:

Spectrum

3 0
27 days ago
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