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2 years ago

6

For the following equilibrium: A+2B⇋C+3D If the change in concentration for B is 0.44 M, what will be the change in concentratio

n for C

1 answer:

Readme [11.4K]2 years ago

5 0

Answer:0.22M

Explanation:

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A. Fossil fuels take a very long time to form so we mine it faster than it can replenish.

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1 year ago

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The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentrati

posledela

Answer:

Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.

Explanation:

When you make a calibration curve in a spectrophotographic analysis you are applying the Lambert-Beer law that states the concentration of a compound is directely proportional to its absorbance:

A = E*l*C

<em>Where A is absorbance, E is molar absorption coefficient, l is optical path length and C is molar concentration</em>

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Using the equation of the line you obtain:

y = 4541.6X + 0.0461

<em>Where Y is absorbance and X is concentration -We will assume concentration is given in molarity-</em>

As absorbance of the unknown is 0.410:

0.410 = 4541.6X + 0.0461

X = 8.01x10⁻⁵M

<h3>Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.</h3>

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2 years ago

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

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2 years ago

Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.

Dimas [21]

Answer:

The concentration of the solution, $C=7.2992\times 10^{-6} M$

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

$A=\varepsilon Cl$

Where,

A is the absorbance of the solution

ɛ is the molar absorption coefficient ( $L.mol^{-1}.cm^{-1}$ )

C is the concentration ( $mol^{-1}.L^{-1}$ )

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 $M^{-1}.cm^{-1}$

l = 1 cm

Applying in the above formula for the calculation of concentration as:

$A=\varepsilon Cl$

$0.20= 27400\times C\times 1$

$C = \frac{0.20}{27400\times 1} M$

So , concentration is:

$C=7.2992\times 10^{-6} M$

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2 years ago

HELP ASAP How many millimeters of mercury pressure does a gas exert at 3.16 atm?

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The conversion factor is 760 mmHg/atm.

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