<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
If Robert has 4 grams of a substance and Jill has 10 grams of the same substance <span>Jill's sample will weigh more than Robert's sample.</span>
Answer:24.31
Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97
Contribution made by isotope of mass 24.99=24.99×10.00=249.9
Contribution made by isotope of mass 25.98=25.98×11.01=286.04
Total contribution=1894.97+249.9+286.04=2430.91
Average mass=2430.91÷100
=24.31
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol