CHEMISTRY COLLEGE

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 35 cm3. If the combustion of this mixture releases 775 J of energy, to what volume will the gases expand against a constant pressure of 710. torr if all the energy of combustion is converted into work to push back the piston

Answers

Answer 1

Answer:

Answer:

The gases will expand 8.2 L against the constant pressure of 710 torr.

Explanation:

Given that:

the original volume V₁ = 35 cm³ = 35 × 10⁻⁶ m³

Since the combustion of the mixture releases energy then :

the work W = - 775 J

Pressure = 710 torr

Since 1 torr = 133.322 Pa

710 torr = 94658.62 Pa

We all know that:

W = -PdV

-775 = - 94658.62 Pa ( V₂ - V₁ )

-775 = - 94658.62 ( V₂ - 35 × 10⁻⁶)

-775/ - 94658.62 = V₂ - 35 × 10⁻⁶

0.008187 = V₂ - 35 × 10⁻⁶

V₂ = 0.008187 + 35 × 10⁻⁶

V₂ = 0.008222 m³

The change in volume dV = V₂ - V₁

The change in volume dV = 0.008222 m³ - 35 × 10⁻⁶ m³

The change in volume dV = 0.008187 m³

To litres

The change in volume dV = 8.2 L

Thus, the gases will expand 8.2 L against the constant pressure of 710 torr.

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Write the full ground state electron configuration of f+.

Answers

Answer:- $1s^22s^22p^4$ .

Explanation:- Atomic number for fluorine(F) is 9 and it's electron configuration is $1s^22s^22p^5$ . $F^+$ is formed when F loses one electron from it's valence shell.

$F\rightarrow F^++1e^-$

Second shell is the valence shell for fluorine and so it loses one electron from 2p to form $F^+$ and the electron configuration of the ion becomes $1s^22s^22p^4$ .

Final answer:

The ground state electron configuration of F+ is 1s²2s²2p⁴, which is derived from the neutral Fluorine's configuration 1s²2s²2p⁵ by removing one electron from the outermost 2p orbital.

Explanation:

The question asked for the full ground state electron configuration of F+. The neutral Fluorine atom (F) has 9 electrons. Its electron configuration is: 1s²2s²2p⁵. When Fluorine loses one electron it becomes a positively charged ion (F+), so its electron configuration will be: 1s²2s²2p⁴. This is because the electron is removed from the outermost shell, specifically the 2p orbital. This ground-state electron configuration refers to the most stable arrangement of electrons around the nucleus.

Learn more about Ground state electron configuration here:

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Which of the following correctly identifies which has the higher first ionization energy, CI or Ar, and supplies the best justification? a. Cl, because of its higher electronegativity b. Cl, because of its higher electron affinity c. Ar, because of its completely filled valence shell d. Ar, because of its higher effective nuclear charge Given the information in a table of bond dissociation energies, calculate the change in enthalpy, delta H, in units of kJ, for the following gas-phase reaction: H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br a. +148 b. -148 c. +200 d. -200

Answers

Answer:

d. Ar, because of its higher effective nuclear charge

For the secon part see explanation below.

Explanation:

The first ionization energy is the energy required to remove an electron from the atom from its outermost shell. It depends on the nuclear charge, distance from the nucleus and the screening of other electrons in the inner shells of the atom.

Comparing Cl and Ar we see that being both elements of the third period, the Ar atom has one more proton than Cl and therefore the electron feels more nuclear charge making the first ionization of Ar greater than Cl.

a) False, electronegativity relates to attraction for an electron and not to the first ionization.

b) False, again electron affinity is not first ionization, it is defined as the energy released when the atom captures an added electron.

c) False,athough it is true that Ar has a complete octet, the higher first ionization is affected by nuclear charge. The screening of electrons in the n= 1 and 2 shells is almost the same so what is important is that the electrons in the n= 3 shell feel more nuclear charge.

d) True for all the reasons given previously : the higher effective nuclear charge in Ar.

For the second part, we have to make an inventory of the bonds being broken and formed:

ΔHºrxn = H broken - H formed, where H is the bond energy

H2 C = CH_2 + H-Br ⇒ CH_3CH_2Br

ΔHºrxn = ( 1 C=C + 4 C-H + 1 H-Br) - ( 1 C-C + 5 C-H + 1 C-Br)

ΔHºrxn (kJ) = (614 + 4(413) + 363) - ( 347 + 5 (413) + 276)

ΔHºrxn (kJ) = 2629 - 2688 = -59 kJ

This value is not in the choices due to mistaken bond energy values from the tables.

Answer:

1. Ar, because of its higher effective nuclear charge.

2. ∆Hrxn = -200 KJ/mol

Explanation:

The size of the atoms of chemical elements can be measured from their atomic radius which is also affected by the effective nuclear charge.

Recall that elements in a particular period have the same number of electron shells. Also, along a given period, atomic radius decreases due to an increase in the effective (positive) nuclear charge. This is because as the atomic (proton) number increases along that period, the charge on the nucleus also increases. With more protons in the nucleus the overall attraction between the positively charged nucleus and the negatively charged surrounding electrons increases, so the electrons are pulled closer to the nucleus thereby leading to a decrease in the atomic size.

So, along a given period atomic size decreases due to an increase in the effective nuclear charge.

The first ionization energy is the minimum energy (in kilojoules) needed to strip one mole of electrons from one mole of a gaseous atom of an element to form one mole of a gaseous unipositive ion.

Along a particular period, ionization energy increases due to an increase in the effective nuclear charge and a decrease in atomic radius. This is because, the smaller the atom the more stable it is and the more difficult it will be to remove an electron.

For the second question,

The enthalpy change of a reaction is the difference in the bond dissociation energies of the reactants and products. Bonds are broken in reactant molecules and formed in product molecules. Bond breaking energies are usually intrinsic ( endothermic, +be ∆H ) while bond forming energies are usually extrinsic ( exothermic, -ve ∆H ).

So,

∆Hrxn = n∆H(reactants/bonds broken) - m∆H(products/bonds formed)

Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation.

First, draw the correct Lewis structures of the compounds.

Next, identify all the bonds broken and formed.

Then, from the bond dissociation energies ( usually given or looked up in texts ), sum up the bond breaking energies and the bond forming energies and subtract the bond forming energies from the bond breaking energies.

Considering this equation:

H_2C = CH_2 + H-Br rightarrow CH_3CH_2Br

The equation is balanced.

Bonds broken (number of bonds ):

I. C=C (1)

II. H-Br (1)

III. C-H (4)

Bonds formed:

I. C-C (1)

II. C-H (5)

III. C-Br (1)

∆Hrxn = [ ( 1 x C=C ) + ( 4 x C-H ) + ( 1 x H-Br ) ] – [ ( 1 x C-C ) + ( 5 x C-H ) + ( 1 x C-Br ) ]

∆Hrxn = [ ( 1 x 614 ) + ( 4 x 413 ) + ( 1 x 141 ) ] – [ ( 1 x 348 ) + ( 5x 413 ) + ( 1 x 194 ) ]

∆Hrxn = [ ( 614+1652+141) ] – [ ( 348 + 2065 + 194 ) ]

∆Hrxn = 2407 – 2607

∆Hrxn = -200KJ/mol

HELP if water vapor condense on a cold surface is the initial and final energy thermal, phase, or chemical?

Answers

My answer to this great question would be thermal

Which of the two aqueous solutions will be a better conductor of electricity in the following examples? Explain why in each case.a)1.0M NaCl vs. 0.010 M Nacl
b)0.10 M NaCl vs. 0.10 M Al2(SO4)3 (assume complete dissociation of both salts).
c) 1.0 mole of NaCl added to water vs. 1.0 mole of pbCl2 added to water (hint: remember the solubility rules).

Answers

a) The solution with a concentration of 1.0 M NaCl will be a better conductor of electricity since it has a higher number of electrolytes
b) The 0.10 M Al2(SO4)3 will be a better conductor since more ions are formed per mole of the salt dissociated
c) 1.0 mol of NaCl in water because NaCl is very soluble in water

What is the value for the kinetic energyfor a n = 2 Bohr orbit electron in Joules?

Answers

Answer:

K.E. = 5.4362 × 10⁻¹⁹ J

Explanation:

The expression for Bohr velocity is:

$v=(Ze^2)/(2 \epsilon_0* n* h)$

Applying values for hydrogen atom,

Z = 1

Mass of the electron ( $m_e$ ) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$v=\frac {2.185* 10^6}{n}\ m/s$

Given, n = 2

So,

$v=\frac {2.185* 10^6}{2}\ m/s$

$v=1.0925* 10^6\ m/s$

Kinetic energy is:

$K.E.=\frac {1}{2}* mv^2$

So,

$K.E.=\frac {1}{2}* 9.1093* 10^(-31)* ({1.0925* 10^6})^2$

K.E. = 5.4362 × 10⁻¹⁹ J

A new process converts potential energy of an object to kinetic energy. Neither work nor eat is added to the system. No reaction takes place. What determines the velocity of the Anew bject? Not sure All of the above Elevation of the process Change in height of the object Mass of the object