Different elements emit different spectra when their electrons get excited because each element has a different arrangement of electrons surrounding the nucleus. The energy levels in which the electrons can occupy are unique to a specific element. When an electron gets excited into a higher energy level, it will eventually relax back into its original state and emit light corresponding to that energy.
276mph that is easy because you times 11.5 with 24 which is the total number of hours per day
The mass of pentane the student should weigh out is
The density of pentane is 0.626 gcm-3
To calculate the mass of pentane following expression is used,
(Density is defined as the mass divide by volume)
Density = mass / volume
mass of pentane = Density of pentane * Volume of pentane
mass of pentane = 0.626 gcm-3 * 45.0 mL
= 28.17 g
Here the unit of mass of pentane is g,
However the unit of density is gcm-3 and unit of volume is mL i.e. cm3
Hence, Mass = gcm-3 * cm3
Mass = g
The mass of pentane the student should weigh out is 28.17g
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Solution:
Benzoic acid is C6H5COOH
In finding pH
C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)
Ka = 6.45 x 10^-6
[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.
Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6
Now,
According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x
x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0
enter a = 1, b = 0.00000645, c = 0.0000323
x = 5.68 x 10^-3 = 0.00568 M expression is [C6H5COOH] = 0.5 M is the correct answer.
[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25
This is the required solution.
