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1 year ago

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A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat

capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl(aq) + NH3(aq) → NH4Cl(aq) The temperature increase is 2.34°C. Calculate ΔH per mole of HCl and NH3 reacted.

1 answer:

Free_Kalibri [48]1 year ago

8 0

Answer:

-154KJ/mol

Explanation:

mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter

= 0.2 × 100 / 1000 ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.

Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K

where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK

heat gain by calorimeter = (4.184 × 200 + 480) × 2.34 = 3081.3 J

ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

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What type of reaction is the digestion of solid copper wire by nitric acid?

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Copper nitrate and nitric oxide are produced in this reaction.

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1 year ago

Tag all the carbon atoms with pi bonds in this molecule. If there are none, please check the box.

snow_tiger [21]

Answer:

Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.

They are frequent components of multiple bonds, as is the molecule indicated in our exercise.

The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.

Explanation:

In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.

Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.

In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.

This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.

When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.

Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.

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2 years ago

6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an

xz_007 [3.2K]

Answer:

ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

Explanation:

1) Reduction potentials

First of all one should look up the reduction potentials for the species envolved:

$Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

$Fe^{3+} + e$ → $Fe^{2+}$ E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>

1) $Ce^{4+}$ reduces while $Fe^{2+}$ oxidates

(oxidation) $Fe^{2+}$ → $Fe^{3+} + e$ E°oxi=-0.771V

(reduction) $Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

(overall equation) $Fe^{2+}+Ce^{4+}$ → $Ce^{3+}+Fe^{3+}$ E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°- $\frac{RT}{nF} Ln Q$

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°= $\frac{RT}{nF} Ln K$

Isolating for 'K' gives:

$K=e^{\frac{nFE^{o} }{RT} }$

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

6 0

2 years ago

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