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3 years ago

| A solution containing 4.48 ppm KMnO4 exhibits

Chemistry

1 answer:

Artist 52 [7]3 years ago

6 0

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = $\frac{4.48 X10^{-3} }{158.03X 1(lit)}$ = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T) ( T = % transmittance)

= - log(0.859)

= 0.06

According to Lambert Beer's law

ε = $\frac{A}{C X l}$

or, ε = $\frac{0.06}{2.83 X 10^{-5}X1 cm }$

or, ε = 2120.14 cm⁻¹M⁻¹

Where

ε = Molar absorptivity

A = absorbance

C = Molar concentration of KMnO₄ solution

l = length

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An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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3 years ago

The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

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3 years ago

One part of the cell theory states that all living things are made up of one or more cells. Scientists had to find ways to test

sergejj [24]

Answer:

A population or community research line can be carried out, wherever at a certain point in time, regardless of whether it is a cross-sectional study.

In addition, the people who would be the population to be studied or the object of study might or might not know the cause of the study (blind) while the researcher could be experimentally participatory.

Explanation:

They are prevalence studies, in which the presence of a health condition or state is determined in a well-defined population and in a determined time frame: one day, one week, a particular moment in life, even if it does not temporarily coincide in all the subjects (for example, the blood pressure figures at the time of entering the school or at the beginning of the holidays, the prevalence of diabetes in hospitalized patients on a given day, etc.).

They are like "photographs" of a state of affairs at a given moment. The simultaneous determination of what is understood by exposure and event does not allow defining causality.

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3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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3 years ago

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2CO + O2 → 2002

Reika [66]

Answer:

B: Increasing the volume inside the reaction chamber

Explanation:

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3 years ago