<h3>
Answer:</h3>
179 L
<h3>
Explanation:</h3>
<u>We are given;</u>
The equation for the combustion of ethane.;
2C₂H₆(g)+7O₂g)→4CO₂(g)+6H₂O(l)
Volume of ethane gas used as 89.5 L
We are required to determine the volume of carbon dioxide produced;
<h3>Step 1: Determine the number of moles of ethane gas</h3>
According to the molar gas volume, 1 mole of a gas occupies a volume of 22.4 L at STP.
Therefore;
Number of moles of ethane = Volume given/molar gas volume
Thus;
Moles of ethane = 89.5 L ÷ 22.4 L
= 3.9955
= 4.0 moles
<h3>Step 2: Determine the number of moles of CO₂ produced </h3>
From the equation,
12 moles of ethane reacts to produce 4 moles of carbon dioxide
Thus; Moles of CO₂ = Moles of ethane × 2
= 4.0 moles × 2
= 8.0 moles
<h3>Step 3: Volume of CO₂ produced</h3>
We know that;
1 mole of a gas = 22.4 L at STP
Therefore;
Volume of CO₂ = 8.0 moles × 22.4 L/mol
= 179.2 L
= 179 L
Therefore, the volume of CO₂ is 179 L
Answer:
4.The elements are organized by their properties.
Explanation:
GASES because they are unseeeable and the electricity cant catch them
The number of moles of NalO3 present in the deposit is 83.23 mol NaIO3.
The mass of NaNO3 deposit present is 2000. Ib Molar mass of NaNO3 is 85.03 g/mol .
Calculate the number of moles of NaNO3
present in the 2000 pounds of NaNO3.
2000 lb x (453.592 gX 1 mol NaNO3)/ 1 lbX 85.03 g
=1.067x10^4 mol NaNO3
The percentage of NaIO3 present in the NaNO3 deposit is 0.78 mol%.
Multiply the number of moles of NaNO3 with the mole percentage of the NaIO3, to get the actual number of moles of NaIO3 present in the deposit.
0.78% (1.067×10* mol NaNO3)=0.78 mol NaIO3/100 mol NaNO,
= 83.23 mol
Thus, the number of moles of NalO3 present in the deposit is 83.23 mol NaIO3
Learn more about NaIO3 here:
brainly.com/question/10249904
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