Answer:
High explosives do not need to be contained to make their bang. Nitroglycerine, trinitrotoluene, and RDX are high explosives.
Explanation:
To calculate the number of molecules, we can simply use this rule:
number of molecules = number of moles x Avogadro's number
In this problem, we have:
number of moles = 4.56 moles
Avogadro's number is a constant = 6.02 x 10^23
Substituting in the above equation, we can calculate the number of molecules as follows:
number of molecules = 4.56 x 6.02 x 10^23
= 2.74512 x 10^24 molecules
The number of Pbr3 that contain 3.68 x10^ 25 bromine atoms is 60.631 moles
calculation
by use of Avogadro law constant
that is, 1 moles = 6.02 x10^23 atoms
what about 3.65 x10^25 atoms
by cross multiplication
= (1 mole x 3.65 x 10 ^25 atoms) / ( 6.02 x10^23 atoms) = 60.631 moles
atoms and atoms cancel out each other
The relation between vapour pressure , enthalpy of vapourisation and temperature is
ln (88/ 39) = DeltaH / 8.314 (1 / 318 - 1 / 298)
0.814 = DeltaH / 8.314 (2.11 X 10^-4 )
DeltaH = -32.07 kJ
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
