The decomposition of formic acid follows first-order kinetics. HCO2H(g) → CO2(g) + H2(g) The half-life for the reaction at 550°C is 24 seconds. How many seconds does it take for the formic acid concentration to decrease by 87.5%? (1) 4.6 seconds (2) 36 seconds (3) 48 seconds (4) 72 seconds (5) 96 seconds

Option 4 is correct (72 seconds)

Explanation:

Option 4 is correct (72 seconds)

The formula we are going to use is:

Where:

A is the final concentration

A_o is the initial concentration

k is the constant

t is the time

Half-Life=0.693/k

Half-life in our case=24 seconds

k=0.693/24

k=0.028875 s^-1

Since the concentration is decreased by 87.5 % which means only 12.5%(100-87.5%) is left.

The ratio will become 0.125

t≅ 72 seconds

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.