Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
You put in a variable to substitute the unknown number.
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
Answer:
Where the greatest electron density is. And isn’t. Look at water. H-O-H.
The oxygen has more electrons, AND 2 lone pairs - the Hydrogens will be partially positive, and so such a way shows the polarity; am arrow with +-> on the Hydrogen area. The “point” will have a negative sign; usually these are drawn as “partially positive” & “partially negative”, but I can’t draw that. Sorry.
Explanation:
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 Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For C and N,
E.N of Nitrogen = 3.04
E.N of Carbon = 2.55
________
E.N Difference 0.49 (Weakly Polar Covalent)
For N and S,
E.N of Nitrogen = 3.04
E.N of Sulfur = 2.58
________
E.N Difference 0.46 (Weakly Polar Covalent)
For K and F,
E.N of Fluorine = 3.98
E.N of Potassium = 0.28
________
E.N Difference 3.70 (Ionic)
For N and N,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference 0 (Non Polar Covalent)
