The number above the symbol is the atomic mass (or atomic weight). This is the total number of protons and neutrons in an atom. The number below the symbol is the atomic number and this reflects the number of protons in the nucleus of each element's atom. Every element has a unique atomic number.
Answer: B) combustion reaction.
Explanation:
A) acid-base reaction: When an acid reacts with a base, to form metal salt and water, this type of reaction is Acid Base reaction.
Example: 
B) combustion reaction: When a hydrocarbon reacts with oxygen to produce carbon dioxide and water, this type of reaction is combustion reaction.
C) precipitation reaction: a reaction in which aqueous solution of two compounds on mixing react to form an insoluble compound which separate out as a solid are called precipitation reactions.
D) gas evolution reaction: a reaction in which one of the product is formed as a gas.
Answer:
Acetylide , Enol ,aldehydes, tautomers, alkynes , Hydroboration, Keto
Explanation:
Reset <u>Acetylide</u> anions are strong nucleophiles that open epoxide rings by an SN2 mechanism. <u>Enol </u>tautomers have an O-H group bonded to a C=C. <u>aldehydes </u>are formed from terminal alkynes with the addition of water using BH3 then H2O2. <u>tautomers</u> are constitutional isomers that differ in the location of a double bond and a hydrogen and exist in an equilibrium with each other. <u>alkynes</u> are compounds that contain a carbon-carbon triple bond. <u>Hydroboration</u> of a terminal alkyne adds BH₂ to the less substituted, terminal carbon.<u> Keto</u> tautomers have a C=O and an additional C-H bond.
Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %
That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.
The given mass of sodium chloride(NaCl) is 45.8 g
Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):
= 1309 g ocean water
Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.
Answer:
First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.
