Well the atomic number for the boron atom is 5
The unit expressed in 660 nm of light represents the wavelength of light. If you want to determine the frequency, you use the speed of light to relate the two. The formula is:
c = λν
where
λ is the wavelength
ν is the frequency
c is the speed of light = 3×10⁸ m
Apply SI units:
(3×10⁸ m) = (660×10⁻⁹ m)(ν)
Solving for ν,
<em>ν = 4.55×10¹⁴ s⁻¹</em>
The osmotic pressure of a solution made from 12.5 g of CaCl2 in enough water to make 500 mL, if CaCl2 is 78.5% dissociated at 30oC is 167.087 atm
<h3>
Define osmotic pressure:-</h3>
Osmotic pressure and osmosis are connected. Osmosis is the passage of a solvent through a semipermeable membrane into a solution. The pressure that halts the osmosis process is known as osmotic pressure. Osmotic pressure is a collective attribute of material because its value is determined by the solute's concentration rather than its chemical composition.
The formula used to calculate Osmotic Pressure is:-
π = M x R x T x i
Where;
π =osmotic pressure (in atm)
M=molar concentration (in mol/L)
=(given mass/molar mass)/volume=(12.5g /111.0)/(500/1000)
=0.224 mol/L
R=Universal gas constant (0.08206 L.atm/mol.K)
T=absolute temperature (in K) = 30.C + 273 = 303 K
i= van't Hoff factor of solute =3
using the formula and substituting values;
π=0.224 x 0.08206 x 303 x 3
= 167.087 atm
The osmotic pressure is 167.087 atm
Learn more about osmotic pressure here:-
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Part 1: Calculate the percent ionization of 0.0075 m butanoic acid (Ka=1.5x10^-5)
C4H8O2(aq) +H2O(l) → C4H7O2(aq) + H3O +
initial 0.0075 0 0
change -X +X +X
final 0.0075-X X X
when Ka is relative smaller to the intial concentration of the acid so we can assume that 0.0075-X≈ 0.0075 by substitution in Ka formula:
Ka = [C4H7O2][H3O+] / [C4H8O2]
1.5x10^-5 = X*X / 0.0075
X^2 = 1.125x10^-7
X= 0.00034 m =3.4 x 10 ^-4 m
∴ [C4H7O2] = [H3O+] = 3.4X10^-4
∴ percent ionization = [H+ equlibrium]/[acid initial] *100
= 3.4X10^-4/0.0075 * 100
= 4.5 %
part 2) calculate the percent ionization of 0.0075m butanoic acid in a solution containing 0.085m sodium butanoic?
C4H8O2(aq) + H2O(l) ↔ C4H7O2(aq) + H3O+
initial 0.0075 0 0.085
change -X +X +X
final 0.0075-X X 0.085+X
we can assume that 0.0075-X≈ 0.0075 & 0.085+X ≈ 0.085
∴Ka = (X*(0.085)) / (0.0075)
(1.5x10^-5)*0.0075 = 0.085X
∴X = 1.3x 10^-6
∴ percent ionization = (1.3x10^-6)/0.0075 * 100 = 0.017 %