T'belongs to the hydrides, it's a nitrogen hydride.
Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
For number 22. The answer is c. To find this look at the numbers before each compound O2 has no number which means 1, while CO2 has a 2 in front
Problem 57EDraw the Lewis structure for CO with an arrow representing the dipole moment. UseFigure 9.10 to estimate the percent ionic character of the CO bond. Step by step solutionStep 1 of 2Lewis structure of CO:Carbon has four valence electrons and oxygen has six valence electrons. If only one bond wereto be formed between carbon and oxygen atoms, carbon would have five electrons and oxygenwould have seven electrons.The single bond between these two atoms is not sufficient to lead to an octet on each atom . Tocomplete the octet of each atom in CO, we must employ a triple bond .A triple bond is formed bythe six electrons sharing between these two atoms and it is shown as follows.\nThe Lewis structure of CO with an arrow representing the dipole moment.The arrow towards anoxygen atom it represents is a highly electronegative atom.
