1,2 I think it’s two tho I’m not sure
Answer:
Explanation:
Given that ;
Compound H is optically active and have a molecular formula of C6H10 and therefore undergo catalytic hydrogenation. Catalytic hydrogenation involves the use Platinum/Nickel to produce C6H12
i.e

The proposed H and I structures are shown in the diagrams attached below .
compound H represents 3- methyl cyclopentene
compound I represents methyl cyclopentane
However; 3- methyl cyclopentene posses just only one chiral carbon which is optically active at the third position and it R and S enantiomers are shown in the second diagram below.
Answer:A 1,2-dibromoheptene
Explanation:
Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M
30 / 1000 L = 0.140 M
0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2
( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol
( 1 L / 0.570 mol )
( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer:
Double replacement or double decomposition
Explanation:
In the reaction below, a double replacement or double decomposition has occurred.
2Na₃PO₄ + 3CaCl₂ → 6NaCl + Ca₃(PO₄)₂
This reaction occurs mainly between two ionic substances.
The reaction involves an actual exchange of partners to form new compounds. For example;
AB + CD → AD + CB
One of the following conditions below serves as the driving force for double replacement reactions:
- Formation of an insoluble compound or precipitate.
- Formation of water or any other non-ionizing compound
- Liberation of a gaseous product.