Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Answer: The formula mass (formula weight) of a molecule is the sum of the atomic weights of the atoms in its empirical formula. The molecular mass (molecular weight) of a molecule is its average mass as calculated by adding together the atomic weights of the atoms in the molecular formula.
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Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
16 g. The mass of 0.60 mol Al is 16 g.
Molar mass of Al = 26.98 g/mol
Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al
