According to the equation you have 1 mole of C2H4 and 3 moles of O2.
1 • (22.4L / 270L) = 3 • (22.4L / x)
1/270L = 3/x
x = 3(270) / 1
x = 810 L
810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP
In an exothermic reaction the energy of the product is less than the energy of the reactants.
Answer : The mass of helium gas added must be 12.48 grams.
Explanation : Given,
Mass of helium (He) gas = 6.24 g
Molar mass of helium = 4 g/mole
First we have to calculate the moles of helium gas.

Now we have to calculate the moles of helium gas at doubled volume.
According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

or,

where,
= initial volume of gas = V
= final volume of gas = 2V
= initial moles of gas = 1.56 mole
= final moles of gas = ?
Now we put all the given values in this formula, we get


Now we have to calculate the mass of helium gas at doubled volume.


Therefore, the mass of helium gas added must be 12.48 grams.
The answer is <span>commensalism</span>
Answer:
first first row: element
second first row :compound