Explanation:
Phosphorus (P) because of the 5 valence electrons total, 3 of them are in the 3p sublevel, and according to Hund's rule, they "single-fill" each orbital first.
The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .
Answer:
Explanation:
Kb of (CH₃)₃N is 7.4 x 10⁻⁵
initial concentration of (CH₃)₃N a is .48 M
(CH₃)₃N + H₂O = (CH₃)₃NH⁺ + OH⁻
a - x x x
x² / (a - x ) = Kb
x is far less than a so a - x can be replaced by a .
x² / a = Kb
x² = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶
x = 5.96 x 10⁻³
pOH = - log ( 5.96 x 10⁻³ )
= 3 - log 5.96
= 3 - .775
= 2.225
A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain
Answer:
[tex]Total = 50.6\ moles[/tex]
Explanation:
Given
[tex]Propane = C_3H_8[/tex]
Represent Carbon with C and Hydrogen with H
[tex]C = 13.8[/tex]
Required
Determine the total moles
First, we need to represent propane as a ratio
[tex]C_3H_8[/tex] implies
[tex]C:H = 3:8[/tex]
So, we're to first solve for H when [tex]C = 13.8[/tex]
Substitute 13.8 for C
[tex]13.8 : H = 3 : 8[/tex]
Convert to fraction
[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]
Cross Multiply
[tex]3 * H = 13.8 * 8[/tex]
[tex]3 H = 110.4[/tex]
Solve for H
[tex]H = 110.4/3[/tex]
[tex]H = 36.8[/tex]
So, when
[tex]C = 13.8[/tex]
[tex]H = 36.8[/tex]
[tex]Total = C + H[/tex]
[tex]Total = 13.8 + 36.8[/tex]
[tex]Total = 50.6\ moles[/tex]
what’s the most abundant isotope of lawrencium
Answer:
266Lr
Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.
Explanation:
hopefully that helps you