For the presence of ammonium ion, there is a need to add sodium hydroxide solution to the water and warm the mixture. Test any vapor that gets produced with damp red litmus paper. It should turn blue as ammonia gas is discharged, which is alkaline. The ionic equation for the reaction is:
NH₄⁺ + OH⁻ ⇒ NH₃ + H₂O
For the presence of phosphate ions, the addition of barium ions is done. The ionic equation is:
3Ba₂⁺ + 2PO4³⁻ ⇒ Ba₃ (PO₄)₂ (precipitate)
Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2<span>,
That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(</span>CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%
You sure the number is correct? Percent yield should not exceed 100%
It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3
density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.
density of object = 2.70 grams / cm^3 <<<<=== answer.
<span>According to Mendeleyev-Klapeyron’s equation
pV = nRT,
where p = 160 atm V = 12 R -constant 0.0821 & T = 298 in Kelvin
Using given data, we can determine the amount of Helium gas:
n = pV/RT = (160â™12)/(0,0821â™298) = 78,48 (mol)
For atmospheric pressure (1 atm) and the same amount we can calculate the volume of tank, using previous equation:
V = nRT/p = (78,48â™0,0821â™298)/1 = 1920 (liters)
V = 1920 liters
Thus Answer is 1920 liters</span>
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>
