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2 years ago

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The molar absorptivity of a compound at 500 nm wavelength is 252 M-1cm-1. Suppose one prepares a solution by dissolving 0.00140

moles of a solute in enough water to make a 500.0 mL solution. What would be the absorbance in a 3 .00 mm pathlength cell?

1 answer:

adell [148]2 years ago

5 0

Answer:

The absorbance of the solution is 0.21168.

Explanation:

Given that,

Wavelength = 500 nm

Molar absorptivity = 252 M⁻¹ cm⁻¹

Number of moles = 0.00140

Volume of solution = 500.0 mL

Length = 3.00 mm

We need to calculate the molar concentration

Using formula of the molar concentration

$C=\dfrac{N}{V}$

Where, N = number of moles

V = volume

Put the value into the formula

$C=\dfrac{0.00140}{0.5000}$

$C=0.0028\ M$

We need to calculate the absorbance of the solution

Using formula of absorbance

$A=\epsilon C l$

Put the value into the formula

$A=252\times0.0028\times0.300$

$A=0.21168$

Hence, The absorbance of the solution is 0.21168.

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2 years ago

A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is

e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

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2 years ago

The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th

kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× $\frac{3molCl_{2}}{2 molCH_{4}}$ = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄× $\frac{3Cl_{2}}{2 molCH_{4}}$ = <em>0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄× $\frac{2CH_{3}Cl}{2 molCH_{4}}$ = <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄× $\frac{2HCl}{2 molCH_{4}}$ = <em>0,0102 mol HCl</em>

0,0102 moles of CH₄× $\frac{2Cl^{-}}{2 molCH_{4}}$ = <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

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I hope it helps!

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2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$

where,

$\Delta G^o$ = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]$

$\Delta G^o=-90.72kJ/mol$

Now we have to calculate the value of $K_p$

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G_^o$ = standard Gibbs free energy = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

$K_p$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p$

$K_p=7.98\times 10^{15}$

Now we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

The expression for equilibrium constant will be :

$K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium $SO_2$ pressure be, x

Pressure of $SO_2$ = Pressure of $H_2S$ = x

Now put all the given values in this expression, we get

$7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}$

$x=3.93\times 10^{-5}torr$

Thus, the equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

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2 years ago

What type of bond joins the carbon atom to each of the hydrogen atoms? A molecule that consists of four hydrogen atoms and one c

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Answer:

Explanation:

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2 years ago