Determine whether each of the following disaccharides is a reducing sugar. Yes, both of the anomeric positions bears an OH group. No, both anomeric positions bear acetal groups. No, one of the anomeric positions bears an OH group. Yes, one of the anomeric positions bears an OH group. No, one of the anomeric positions bears an OH group. No, both anomeric positions bear acetal groups. Yes, one of the anomeric positions bears an OH group. Yes, both of the anomeric positions bears an OH group. No, both anomeric positions bear acetal groups. Yes, one of the anomeric positions bears an OH group. Yes, both of the anomeric positions bears an OH group. No, one of the anomeric positions bears an OH group.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

mark him as !

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :