Answer:
1, 1, 2
Explanation:
You need to have the same amount of atoms of everything element on both the products and reactants side
Reactants:
Na=2
O=2
H=2
Products:
Na=2
O=2
H=2
A. Barium with 56 is correct
The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.
b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.
c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.
Question:
What type of product forms in the intramolecular reaction between the aldehyde portion of the glucose molecule below and its C-5 hydroxyl group?
a. disaccharide
b. carboxylic acid
c. hemiacetal
d. ester
e. stereoisomer
Answer:
hemiacetal forms in the intramolecular reaction between the aldehyde portion of the glucose molecule and its C-5 hydroxyl group
Explanation:
It is an alcohol also an ether that has been attached to the carbon molecule. Here the hydrogen has occupied the fourth bonding position. This hemiacetal has been derived from the aldehyde. Hence, hemiketal being an alcohol as well as ether has been attached to the same carbon and also to the two other carbon.