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2 years ago

What's the principle of Atomic emission spectroscopy?

Chemistry

1 answer:

slega [8]2 years ago

4 0

Answer:

Explanation:

The theory or working principle of Atomic Emission Spectroscopy involves the examination of the wavelengths of photons discharged by atoms and molecules as they transit from a high energy state to a low energy state. A characteristic set of wavelengths is emitted by each element or substance which depends on its electronic structure.

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3. Methyl acetate is hydrolyzed at 25 oC in acidic environment. Aliquots of equal volume are removed and titrated with NaOH solu

avanturin [10]

Solution :

Time (sec) Volume of NaOH (mL)

339 26.23

1242 27.80

2745 29.70

4546 3.81

$\infty$ 39.81

Now the example of the first order kinetics w.r.t volumetric analysis is :

$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$

Here, $v_{\infty}= \text{ volume at }\infty = 39.81$

$v_{t}= \text{ volume at time 't' } = 27.80$

$v_0$ = volume at time 0 = 0

Since the interval is not constant, we take the time interval as

$=\frac{903+1503+1801}{3}$

$=\frac{4207}{3}$

= 1402.3333

≈ 1402 seconds

$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$

$=(0.001643) \log \left(\frac{39.81}{10.01}\right)$

= 0.001643 x 0.52045

= 0.00082

$= 8.55 \times 10^{-4} \ sec^{-1}$

Therefore, the first order rate constant is k $= 8.55 \times 10^{-4} \ sec^{-1}$ .

5 0

2 years ago

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

Answer: The above reaction is non-spontaneous.

Explanation:

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0

3 years ago

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is

Tanzania [10]

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

Full question is:

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

4 0

3 years ago

A gaseous substance turns into a solid. Which best describes this change? A substance that has a specific shape changes to a sub

Snezhnost [94]

A substance that has no specific volume changes to a substance that has a specific volume.

8 0

3 years ago

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