Draw the product formed when cyclohepta-1,3,5-triene (pka = 39) is treated with a strong base.

you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only. therefore, you had 15.67 7.58 = 8.09g h2o moles of each : 7.58g mgco3 = moles = 7.58g / 83.4g/mole = 0.09moles 8 09g h2o = moles = 8.09g / 18g/mole = 0.44moles molar ratio between the two by dividing both numbers by the smallest number : 0.09 / 0.09 = 1 0.44 / 0.09 = 4.8 mgco3-5h2o

you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only. therefore, you had 15.67 7.58 = 8.09g h2o moles of each : 7.58g mgco3 = moles = 7.58g / 83.4g/mole = 0.09moles 8 09g h2o = moles = 8.09g / 18g/mole = 0.44moles molar ratio between the two by dividing both numbers by the smallest number : 0.09 / 0.09 = 1 0.44 / 0.09 = 4.8 mgco3-5h2o

you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only. therefore, you had 15.67 7.58 = 8.09g h2o moles of each : 7.58g mgco3 = moles = 7.58g / 83.4g/mole = 0.09moles 8 09g h2o = moles = 8.09g / 18g/mole = 0.44moles molar ratio between the two by dividing both numbers by the smallest number : 0.09 / 0.09 = 1 0.44 / 0.09 = 4.8 mgco3-5h2o

This post is last updated on hrtanswers.com at Date : 1st of September – 2022