Answer:
23.0733 L
Explanation:
The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:
Mass = 62.5 g
Molar mass of = 34 g/mol
The formula for the calculation of moles is shown below:
Thus, moles are:
Consider the given reaction as:
2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.
Also,
1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.
So,
1.8382 moles of hydrogen peroxide decomposes to give
So,
Pressure = 746 / 760 atm = 0.9816 atm
Temperature = 27 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (27 + 273.15) K = 300.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K
<u>⇒V = 23.0733 L</u>
Answer:
hi buddy!! here is ur ans: carbon atoms
Answer:
The mass of water produced is 4.29 g.
Explanation:
Given;
reacting mass of ethane, C₂H₆ = 2.5 g
reacting mass of oxygen, O₂ = 5.0 g
The balanced combustion reaction is given as follows;
2C₂H₆ + 7O₂ ----------------> 4CO₂ + 6H₂O
Based on the balanced equation above;
requires
7.0 moles of oxygen -----------------------> 2.0 moles of ethane
requires
5.0 g of oxygen --------------------------> x gram of ethane
x = (2 x 5) / 7
x = 1.43 g
Ethane is in excess of 1.07g (2.5 g - 1.43g)
Now, determine the mass of water produced by 1.43 g of ethane;
requires
2.0 moles of ethane -----------------------> 6.0 moles of water
requires
1.43 g of ethane --------------------------> y gram of water
y = (6 x 1.43) / 2
y = 4.29 g
Therefore, the mass of water produced is 4.29 g.
"<span>4 Na + O2 2 Na2O" is the one chemical equation among the following choices given in the question that </span><span>is correctly balanced. The correct option among all the options that are given in the question is the fourth option or option "4". I hope that this is the answer that has actually come to your great help.</span>