Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:

= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
Answer:
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Explanation:
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Answer: It will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s
Explanation:
Zero order reaction : A reaction is said to be of zero order if the rate is independent of the concentration of the reactants, that means the rate is directly proportional to the zeroth power of the concentration of the reactants.
Expression for the zero order kinetics:
![[A]=-kt+[A]_o](https://tex.z-dn.net/?f=%5BA%5D%3D-kt%2B%5BA%5D_o)
where [A] = concentration left after time t = 0.210 M
= initial concentration = 0.900 M
k= rate constant =
t = time for reaction = ?


Thus it will take for the concentration of A to decrease from 0.900 M to 0.210 M is 28.75 s
Answer:
9.15×10²³molecules
Explanation:
moles=number of particles/Avogadro's number
1.52=x/6.02×10²³
by cross multiplication;
x=1.52×6.02×10²³
=9.15×10²³
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Answer:
The sum of all the charges should equal the charge on the ion.