Rydberg Eqn is given as:
1/λ = R [1/n1^2 - 1/n2^2]
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
w/w percentage <span>
= mass of the pure compound /
total mass of the sample x 100%
70% HNO₃
contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>
The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
= 103.8 g x (70 / 100)
<span>
= 72.66 g = 72.7 g.</span></span>
Answer: Mixture: Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze
Pure substance: Copper wire, distilled water, table salt, oxygen.
Explanation:
Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.
Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze
Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.
Copper wire, distilled water, table salt, oxygen.
Answer:
Drug calculation
If we have 45g of clobetasol = 0.05%w/w
Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g
It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w
Answer:
four (4)
Explanation:
Naphthalein is an organic compound with formula C
10H
8. It is the simplest polycyclic aromatic hydrocarbon, and is a white crystalline solid with a characteristic odor that is detectable at concentrations as low as 0.08 ppm by mass. As an aromatic hydrocarbon, naphthalene's structure consists of a fused pair of benzene rings. It is best known as the main ingredient of traditional mothballs.
The molecule is planar, like benzene. Unlike benzene, the carbon–carbon bonds in naphthalene are not of the same length. The bonds C1−C2, C3−C4, C5−C6 and C7−C8 are about 1.37 Å (137 pm) in length, whereas the other carbon–carbon bonds are about 1.42 Å (142 pm) long. This difference, established by X-ray diffraction is consistent with the valence bond model in naphthalene and in particular, with the theorem of cross-conjugation. This theorem would describe naphthalene as an aromatic benzene unit bonded to a diene but not extensively conjugated to it (at least in the ground state), which is consistent with two of its three resonance structures.
Because of this resonance, the molecule has bilateral symmetry across the plane of the shared carbon pair, as well as across the plane that bisects bonds C2-C3 and C6-C7, and across the plane of the carbon atoms. Thus there are two sets of equivalent hydrogen atoms: the alpha positions, numbered 1, 4, 5, and 8, and the beta positions, 2, 3, 6, and 7. Two isomers are then possible for mono-substituted naphthalenes, corresponding to substitution at an alpha or beta position. Bicyclo[6.2.0]decapentaene is a structural isomer with a fused 4–8 ring system.
Therefore four (4) double bonds will be added to give each carbon atom an octet structure.