"The Oak Tree"

Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the

Alecsey [184]

Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

$\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

102.5 = $(\frac{1}{2})x + 498 - 486$

102.5 - 12 = $\frac{x}{2}$

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

$\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

= 339.5 - 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

8 0

3 years ago

1 point Berny has a 0.15-kilogram candle, a 1.00-kilogram jar, and a 0.25-kilogram candle holder. He places the candle and candl

evablogger [386]

Answer:

1.40

Explanation:

Study Island

5 0

3 years ago

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The

dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

$_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e$

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

$_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e$

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0

3 years ago

A sample of o2 gas (2.0 mmol) effused through a pinhole in 5.0 s. it will take __________ s for the same amount of co2 to effuse

s344n2d4d5 [400]

Answer is: 5,9.
Effusion is leakage of gas through a small hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
</span>M(O₂) = 32g/mol.
M(CO₂) = 44g/mol.
t₁ = 5s.
t₂ = ?
t₂/t₁ = √(M(CO₂)/M(O₂)).
t₂ = 1,17·5s = 5,9s.

8 0

3 years ago

Read 2 more answers

The stability of an isotope is based on its

sattari [20]

Option 4. ratio of electrons to protons

Isotope that has atomic number of 82 is stable. An element that has an atomic number ∠82 more stable except Tc and Pm. Also there is the concept that isotopes consisting a combination of even-even,even-odd,odd-even, and odd-odd are all stable. Many isotopes with no magic numbers of nucleons are stable

7 0

3 years ago