At 50 degrees Celsius and standard pressure inter-molecular forces of attraction are strongest in a sample of ethanoic acid.
Ethanoic acid has hydrogen atom bonded with a more electronegative atom; Oxygen. As a result, the molecule possesses strong intermolecular Hydrogen Bonds. Therefore; ethanoic acid, and all other carboxyllic acids have the tendency to form dimers.
Answer:
H₃PO₄ → 3H⁺ + PO₄³⁻
CaSO₄ → Ca²⁺ + SO₄²⁻
b. CaCl₂
Explanation:
When H₃PO₄ is dissolved in water, there are produced the H⁺ and PO₄³⁻ ions. The equation is:
H₃PO₄ → 3H⁺ + PO₄³⁻
In the same way, CaSO₄ is dissolved in:
CaSO₄ → Ca²⁺ + SO₄²⁻
b. Now, in a reaction of an acid (HCl) and a base (Ca(OH)₂), water, H₂O and a salt are produced:
2 HCl + Ca(OH)₂ → 2H₂O + Salt
The ions that are not present in the reaction are Cl⁻ and Ca²⁺, the salt is CaCl₂ and the balanced reaction is:
2 HCl + Ca(OH)₂ → 2H₂O + CaCl₂
Answer:
a) 
b) 
c) 
Explanation:
The symbols of the isotopes are written like
where,
X is the element
A is the mass number (protons + neutrons)
Z is the atomic number (protons)
<em>a) Iodine-131</em>
The atomic number of iodine is 53. The mass number of this isotope is 131. The symbol is .
<em>b) Iridium-192</em>
The atomic number of iridium is 77. The mass number of this isotope is 192. The symbol is .
<em>c) Samarium-153</em>
The atomic number of samarium is 62. The mass number of this isotope is 153. The symbol is .
Answer:
-255.4 kJ
Explanation:
The free energy of a reversible reaction can be calculated by:
ΔG = (ΔG° + RTlnQ)*n
Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.
C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
Q = pC₂H₆/[pC₂H₂ * (pH₂)²]
Q = 0.261/[8.58*(3.06)²]
Q = 3.2487x10⁻³
ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)
ΔG = -255.4 kJ
