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Kruka [31]
3 years ago
7

An aqueous solution is 1.00% by mass ethanol, CH3CH2OH, and has a density of 0.996 g/mL.

Chemistry
1 answer:
Katarina [22]3 years ago
3 0

The molarity of ethanol in the solution is 0.217 M

<h3>Further explanation</h3>

Given

1% by mass

density = 0.996 g/ml

Required

The molarity

Solution

mass/ml of Ethanol :

1% x 0.996 g/ml = 0.00996 g/ml

For 1 L solution :

0.00996 g/ml x 1000 ml/L = 9.96 g/L

Convert to mol/L(MW ethanol=46 g/mol) :

9.96 g/L : 46 g/mol= 0.217 mol/L = 0.217 M

Or you can use equation :

\tt M=\dfrac{\%mass\times \rho\times 10}{MW}\\\\M=\dfrac{1\times 0.996\times 10}{46}=0.217

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At what temperature is the following reaction feasible: HCl(g) + NH3(g) -&gt; NH4Cl(s)?
Nutka1998 [239]
Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).

All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot must be >=0 for a chemical change to be feasible.

For example: CaCO3(s) ==> CaO(s) + CO2(g) 

ΔSθsys = ΣSθproducts – ΣSθreactants 

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),

Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.

But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys +  ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

8 0
3 years ago
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Calculate the volume of an object that has a density of 4 g/mL and has a mass of 128 grams. Show your work
SOVA2 [1]
The first one is 32mL and the second one is 2.62 and I think it’s grams/mL I’m not for sure about the letters on the second one

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3 years ago
Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.
Cerrena [4.2K]

Answer:

Answer :

The total pressure inside the cylinder is, 6.7 atm

The mole fraction of N_2 in the mixture is, 0.52

Solution :

First we have to calculate the total pressure inside the cylinder.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

P_T=p_{N_2}+p_{O_2}+p_{Ar}+p_{He}

Now put all the given values is expression, we get the total pressure inside the cylinder.

P_T=3.5+2.8+0.25+0.15=6.7atm

Now we have to calculate the mole fraction of N_2 in the mixture.

Formula used :

pN_2=XN_2 x P_T

where,

P_T = total pressure = 6.7 atm

pN_2 = partial pressure of nitrogen gas = 3.5 atm

XN_2 = mole fraction of nitrogen gas = ?

Now put all the given values in the above formula, we get

3.5atm=XN_2 x 6.7atm

XN_2=0.52

Therefore, the total pressure inside the cylinder is, 6.7 atm and the mole fraction of N_2 in the mixture is, 0.52

Hope it helps answer the question:)

5 0
3 years ago
A particular graduated cylinder contains 24.0 mL Br2(liquid). The density of bromine at 25 degrees C is 3.12g/cm^3. Note that in
motikmotik

Answer:

2.82x10^{23}molecules

Explanation:

Hello.

In this case, given the volume (1cm³=1mL) and density of the bromine we are to firstly compute the mass since it will allow us to compute the representative particles:

\rho =\frac{m}{V}

m=\rho *V=3.12g/cm^3*24.0cm^3\\\\m=74.88gBr_2

Next, since the mass of one mole of diatomic bromine is 159.82 g (one bromine weights 78.91), we can next compute the moles in that sample:

n=74.88g*\frac{1mol}{159.82g} =0.469molBr_2

Finally, via the Avogadro's number we can compute the representative particles of bromine as follows:

particles=0.469mol*\frac{6.022x10^{23}molecules}{1mol}\\ \\2.82x10^{23}molecules

Best regards.

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3 years ago
What do a mole of magnesium (Mg) and a mole of iron (Fe) have in common?
gavmur [86]

Answer:

A. Their numbers of atoms

Explanation:

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