Answer:
of water at 30C and 1 atm is 256.834 J/mol·K.
Explanation:
To solve the question, we note the Maxwell relation such as
Where:
= Specific heat of gas at constant pressure = 75.3 J/mol·K
= Specific heat of gas at constant volume = Required
T = Temperature = 30 °C = 303.15 K
α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹
K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹
Therefore,
75.3 - 
= 
= - 75.3 = 256.834 J/mol·K.
Answer:
3.6 × 10²⁴ molecules
Explanation:
Step 1: Given data
Moles of methane (n): 6.0 moles
Step 2: Calculate the number of molecules of methane in 6.0 moles of methane
In order to convert moles to molecules, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ molecules of methane in 1 mole of molecules of methane.
6.0 mol × 6.02 × 10²³ molecules/1 mol = 3.6 × 10²⁴ molecules
Answer:
bdndbdjdbdjdbdjdbsidbdidbsjsbsisbsidbd
Answer:
I think
Explanation:
low charge of ions, large cation and small anion
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)
