A boy lifts a 30N dragon 2 meters above the ground. How much work did the boy do on the dragon?

Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t

Contact [7]

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)Δ $H^{0}f(products)$ - ∑n(reactants)Δ $H^{0}f(reactants)$

where n = number of moles

Δ $H^{0}f$ = enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

3 0

2 years ago

Which neutral atom is isoelectronic with o+?

sukhopar [10]

Isoelectronic means equal number of electrons.

O+ is formed when the atom of O loses 1 electron.

The number of electrons of neutral O atom equals its number of protons.

Number of protons identifies the atomic number and position of the element in the periodic table.

The positon of O in the periodic table is A = 8, so it has 8 electrons and O+ has 8 - 1 = 7 electrons.

The neutral atom with one electron less than O is of the element to the left of O in the periodic table (A = 7). That element is N.

Therefore, the neutral atom isoelectronic with O+ is N (both have 7 electrons).

5 0

2 years ago

Gold is one of the densest substances known, with a density of 19.3 g/cm3. If the gold in the crown was mixed with a less-valuab

Andrews [41]

Answer:

Explanation:

Density of gold is 19.3 g / cm³

Density of copper is 8.96 g / cm³

Density of bronze is 8.7 g / cm³

Hence when the gold and copper or bronze are mixed , the density of gold will be reduced due to less density of copper and bronze in comparison to that of gold.

4 0

2 years ago

Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which

VMariaS [17]

Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) $sp^{3}$ hybrid orbitals invovle sigma bonding.

$\pi$ orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

a)

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

b)

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is $sp^{3}$

c)

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves $\pi$ bonding.

Therefore, $\pi$ bonds are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

5 0

2 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$