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3 years ago

Contrast the bonds that are formed from diatomic chlorine and diatomic oxygen.

Chemistry

1 answer:

melomori [17]3 years ago

6 0

<h3>Answer;</h3>

Chlorine forms one covalent bond and oxygen forms two covalent bonds

<h3>Explanation;</h3>

Covalent bond is a type of bond the occur when non-metal atoms share electrons. Covalent atom may occur between two atoms in a molecule.
Diatomic molecule is a molecule consisting of two atoms. Halogens such as chlorine exist as diatomic gases by forming covalent bonds.
Single covalent bond occurs when two atoms held together by sharing a pair of electrons, for example in diatomic molecules such as chlorine.
Double covalent bond is a bond that involves two shared pairs of electrons for example in oxygen molecule. Triple covalent bond on the other hand is a bond formed by sharing three pairs of electrons between atoms, like in the case of nitrogen molecule.

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A sample of oxygen occupies a volume of 150 cm3 at 72 C. What will be volume of oxygen when the temperature drops to 2 C?

Ksenya-84 [330]

Answer:

V₂ = 119.59 cm³

Explanation:

Given data:

Initial volume = 150 cm³

Initial temperature = 72°C

Final volume = ?

Final temperature = 2°C

Solution:

Initial temperature = 72°C (72 +273.15 K = 345.15 K)

Final temperature = 2°C (2 + 273.15 k = 275.15 K)

Solution:

According to the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 cm³ × 275.15 K /345.15 K

V₂ = 41272.5 cm³.K / 298 K

V₂ = 119.59 cm³

8 0

2 years ago

Read 2 more answers

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O

trasher [3.6K]

Answer: $\Delta H^{0} = -879.15 kJ/mol$

Explanation: Heats of formation is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

Standard Enthalpy Change is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

$\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}$

For the reaction

2NH₃ + 3N₂O → 4N₂ + 3H₂O

2(-46.2) + 3(82.05) 4(0) + 3(-241.8)

$\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]$

$\Delta H^{0}=-725.4-153.75$

$\Delta H^{0}=-879.15$

The standard enthalpy change for the reaction is $\Delta H^{0}=-879.15$ kJ

8 0

2 years ago

Suggest 2 reasons why the noble gases become less ideal in their behaviour down the group from helium to xenon.

Elenna [48]

Reason1: electrons on farther layers become free easyer
2nd reason: volume of atoms grows (from Helium to Xeon) so instead of boyle-mariot law equation (PV=vRT) is more accurate to use van der walls equations that adds to the boyle-mariot equation the volume occupied by the atoms of the gas to the volume of the space between the atoms P(Vm-b)=vRT

4 0

3 years ago

A beaker of water has a volume of 125mL and a density of 1.0g/mL. Calculate the mass of the water.

Oksanka [162]

Answer: 125 grams

Explanation: (125mL)*(1.0g/mL) = 125g

4 0

2 years ago

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

3 years ago