Answer:
<u></u>
Explanation:
Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.
<u>1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution</u>
- number of moles = molarity × volume in liters
- number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol
<u>2. Determine the number of moles of sulfuric acid needed</u>
- number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol
<u>3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.</u>
- Molarity = number of moles / volume in liters
- M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M
Round to two significant figures: 0.37M
Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:

We have :
Enthalpy changes of formation of following s:



(standard state)
![\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5B%5CDelta%20H_f%28product%29%5D-%5Csum%20%5B%5CDelta%20H_f%28reactant%29%5D)
The equation for the enthalpy change of the given reaction is:
=

=


The enthalpy change for the given reaction is 424 kJ.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.