I need help solving this chemistry question

The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO

Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy = 435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435 + 431 + x

x = -92 - (435 + 431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

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6 0

1 year ago

Suppose you have 100 grams of a radioisotope with a half-life of 100 years. How much of the isotope will you have after 200 year

Ulleksa [173]

Amount Remaining Years #half lives
100g 0 0
50 g 100 1
25g 200 2

8 0

3 years ago

Read 2 more answers

With white light, a team measured a 0.7% percent change with 3.5g of plant matter in a one liter container. Convert

Strike441 [17]

moles CO₂ = 5.57.10⁻⁴

<h3>Further explanation </h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

Can be formulated :

$\tt mol=\dfrac{mass}{MW}$

0.7% percent change with 3.5g of plant matter

mass :

$\tt 0.7\%\times 3.5~g=0.0245~g$

moles :

$\tt moles=\dfrac{0.0245}{44}=0.000557=5.57.10^{-4}$

5 0

3 years ago

For each of the following sublevels, give the n and l values and the number of orbitals: (a) 6g; (b) 4s; (c) 3d.

olya-2409 [2.1K]

Answer:

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 5

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

(c) 3d. Shell 3, n = 3. Subshell d, l = 2. Number of orbitals in sublevel = 2l+1 = 5

4 0

3 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0

3 years ago

I need help solving this chemistry question ​

I need help solving this chemistry question