In an open flame because it can not only catch on fire extremely easily but also cause an explosive
<span>Carbon Monoxide.
First, determine the relative number of moles of each element by looking up the atomic weights of carbon and oxygen
Atomic weight carbon = 12.0107
Atomic weight oxygen = 15.999
Moles of Carbon = 24.50 g / 12.0107 g/mol = 2.039847802 mol
Moles of Oxygen = 32.59 g / 15.999 g/mol = 2.037002313 mol
Given that the number of moles of both carbon and oxygen are nearly identical, it wouldn't be unreasonable to think that the empirical formula for the compound is CO which also happens to be the formula for Carbon Monoxide.</span>
Answer: 625 kj/mol
Explanation:
As shown below this expression gives the activation energy of the reverse reaction:
EA reverse reaction = EA forward reaction + | enthalpy change |
1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:
EA = energy of the transition state - energy of the reactants.
2) The activation energy of the forward reaction given is:
EA = energy of the transition state - energy of [ NO2(g) + CO(g) ] = 75 kj/mol
3) The negative enthalpy change - 250 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 375 kj / mol - 250 kj / mol = 125 kj/mol
4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.
5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:
EA reverse reaction = EA forward reaction + | enthalpy change |
EA reverse reaction = 375 kj / mol + |-250 kj/mol | = 375 kj/mol + 250 kj/mol = 625 kj/mol.
And that is the answer, 625 kj/mol
Answer:
Explanation:
1. Solubility of CaF_2
(a) Molar solubility
CaF₂ ⇌ Ca²⁺ + 2F⁻
![K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bsp%20%7D%7D%20%3D%20%5Ctext%7B%5BCa%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BF%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%282s%29%5E%7B2%7D%3D4.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5C4s%5E%7B3%7D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%5E%7B3%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%20%3D2.2%20%5Ctimes%2010%5E%7B-3%7D%5Ctext%7B%20mol%2FL%7D)
(b) Mass solubility
2. pH
pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52
3. Oxidizing and reducing agents
Zn + Cl₂ ⟶ ZnCl₂
The oxidation number of Cl has decreased from 0 to -1.
Cl has been reduced, so Cl is the oxidizing agent.
4. Oxidation numbers
(a) Al₂O₃
1O = -2; 3O = -6; 2Al = +6; 1Al = +3
(b) XeF₄
1F = -1; 4F = -4; 1 Xe = +4
(c) K₂Cr₂O₇
1K = +1; 2K = +2; 1O = -2; 7O = -14
+2 - 14 = -12
2Cr = + 12; 1 Cr = +6
Answer:
Due to presence of a triple bond between the two N−atoms, the bond dissociation enthalpy (941.4 kJ mol
−1
) is very high. Hence, N
2
is the least reactive.
