Answer:
V = 10.3 L
Explanation:
Given data:
Mass of methane = 6.40 g
Volume of CO₂ produced = ?
Temperature = 35°C (35+273 = 308 K)
Pressure = 100.0 KPa (100.0/101 = 0.98 atm)
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 6.40 g/ 16 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of CO₂ with CH₄.
CH₄ : CO₂
1 : 1
0.4 : 0.4
Volume of CO₂:
Formula:
PV = nRT
0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K
0.98 atm ×V = 10.11 atm.L
V = 10.11 atm.L /0.98 atm
V = 10.3 L
In the phase diagram the line between solid and liquid is where both phases available at the same time
Answer is (b)
2,3,1 is the stoichiometry
The values in front of the elements are the stoichiometric values
(Since Al2Br6 has no value in front, it's considered 1)
iondine
Explanation:
bromine has 185pm, astatine haa 200, iondine has 140 and lastly fluorine has 147. so iondine has the smallest atomic radius
C) Noble gases
The six noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). Their atomic numbers are, respectively, 2, 10, 18, 36, 54, and 86.
