1. B
The positive charge in water is provided by hydrogen, and gold provides the same charge. However, gold is not more reactive than hydrogen so it can not replace it in the compound.
2. In order to balance the equation, you must sure there are equal moles of each element on the left and right side of the equation:
2C₂H₆ + 7O₂ → 4CO₂ + ₆H₂O
3. The number of moles of sodium atoms on the left of the equation must be equal to the number of moles of sodium atoms on the right, as per the law of conservation of mass. The answer is B.
4. C.
A synthesis reaction usually results from single displacement because some element or compound is produced in its pure form
5. B.
The gas being produced is being synthesized.
Answer :
- Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as
Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.
The general formula of carboxylic is, 
. According to the IUPAC naming, the carboxylic are named as alkanoic acids.
Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.
The general representation of aldehyde is, 
. According to the IUPAC naming, the aldehyde are named as alkanals.
Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.
The general representation of ketone is, 
. According to the IUPAC naming, the ketone are named as alkanone.
Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.
The general representation of ester is, 
. According to the IUPAC naming, the ester are named as alkyl alkanoate.
Concentration is the number of moles of solute in a fixed volume of solution
Concentration(c) = number of moles of solute(n) / volume of solution (v)
25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.
original solution molarity - 0.150 M
number of moles of LiOH in 1 L - 0.150 mol
number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol
when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases
new volume - 125 mL + 25 mL = 150 mL
therefore new molarity is
c = 0.01875 mol / 0.150 L = 0.125 M
answer is 0.125 M
I think 1.00 mol sorry if I’m wrong
