Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
Positive because it keeps going ok
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =
Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =
Thus vant hoff factor for sodium chloride in X is 1.9
Group 18 is the only group in periodic table in which all the elements are nonmetals. This group contains F, Cl, Br, I and At and also the other name of this group is halogen which means salt producer.
You welcome and please give me brainiest!
Answer:
4Fe + 3O₂ → 2Fe₂O₃
General Formulas and Concepts:
<u>Chemistry - Reactions</u>
Explanation:
<u>Step 1: Define</u>
RxN: Fe + O₂ → Fe₂O₃
<u>Step 2: Balance</u>
We need to balance both Fe and O.
LCM of 2 and 3 is 6:
Fe + 3O₂ → 2Fe₂O₃
We now need the same amount of Fe on both sides:
4Fe + 3O₂ → 2Fe₂O₃