The whole Activity , poem and paragraph is missing in the question.
Answer:
(1) Liquid A
(2) Solid A
Explanation:
Using this part of the given poem
Substances and mixtures behave differently,
During boiling and melting most especially
Boiling point of substance is fixed while mixture is not
Substance melts completely but mixture does not
The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A
Solid A is melting completely so Solid A is a pure substance.
Answer:
The percent yield of this reaction is 70%
Explanation:
The reaction is: N₂ + 3H₂ → 2NH₃
We only have the mass of H₂, so we assume that N₂ is in excess
We convert the mass to moles, to work with the reaction:
450 g . 1mol / 2 g = 225 moles
Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia
Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles
This is the 100% yield reaction → We convert the moles of NH₃ to mass
150 mol . 17g /1mol = 2550 g
Percent yield = (Produced yield/Theoretical yield) .100
Percent yield = (1575g/2550g) . 100 = 70%
Answer:
1 electron is involved.
Explanation:
Hello,
In redox reactions, when therer's the necessity to know the involved equivalents, they equal the number of transferred electrons, in this case, since one equivalent is stated, one electron is transferred (involved).
Best regards.
Answer:
Nsc=30
Explanation:
The solid angle subtended by the counter is
d=0.1mm2/(1cm)2=(10)3 sr
d=(10)10x10.5gcm3x10-4cm/108x1.6610x10-24gx0.510-24cm2x10-3=30
Hybridization in ozone, O3......
<span>...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O </span>
<span>../ </span>
<span>O .................. 3 lone pairs on terminal O </span>
<span>I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. </span>
<span>The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. </span>
<span>If the electron pair geometry is linear, the hybridization is sp. </span>
<span>If the electron pair geometry is trigonal planar, the hybridization is sp2. </span>
<span>If the electron pair geometry is tetrahedral, the hybridization is sp3. </span>
<span>The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.</span>
