Answer : A - carboxyl group.
The carbon in the carboxyl group is bonded to a hydroxyl group and also it is double bonded to an oxygen group.
The carboxyl group when it is dissolved in a solution donates it's hydrogen ion(H+ ion). When it is dissolved in water, the carboxyl group loses its H+ ion to form negatively charged ion.
RCOOH + H2O⇄RCOO- + H3O+
Thus the carboxyl group act as an acid when dissolved in water.
The only thing you need to put the followings into correct order is to remember that the smalles mass - the less volume it has. So, accordnig to this, the order should be like that : An ice crystal- A water molecule-An oxygen atom-The electron cloud-<span>An oxygen nucleus. Hope you will find it helpful!</span>
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
The electrons will move more rapidly resulting in a higher pressure even at a consistent volume
