For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
Oxygen is limiting reactant
Explanation:
Based on the chemical reaction:
2C2H6 + 7O2 → 6H2O + 4CO2
<em>2 mole of ethane reacts with 7 moles of oxygen</em>
<em />
For a complete reaction of 5.25 moles of ethane are required:
5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen
As there are just 15.0 moles of oxygen
<h3>Oxygen is limiting reactant</h3>
The molar concentration is 1.11M.
<h3>What is molar concentration?</h3>
The phrase "molar concentration" (also known as "molarity," "amount concentration," or "substance concentration") refers to the amount of a substance per unit volume of solution and is used to describe the concentration of a chemical species, specifically a solute, in a solution. The most frequent measure of molarity in chemistry is the number of moles per liter, denoted by the unit symbol mol/L or mol/dm3 in SI units. A solution with a concentration of 1 mol/L is referred to as 1 molar, or 1 M.
<h3>Given : </h3>
Volume of the solution = 2L
Mass of glucose given = 200g
Concentration of glucose= ?
<h3>Formula use: </h3>
Molarity = no. of moles of solute / volume of the solution (L)
Moles of solute = given mass of solute / molar mass of the solute
<h3>Solution: </h3>
No. of moles of solute( glucose ) = 200 / 180 = 1.11 moles'
Molarity = 1.11 / 2 = 0.5555 mol L ^(-1)
Therefore, the molar concentration of glucose in the solution = 0.555 mol L ^(-1)
To learn more about molar concentration :
brainly.com/question/15532279
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Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= 
= 
= 
The amount of free SO₂ to be added will be:
= 
= 
∵ 1000 mg = 1 g
So,
= 
= 
Thus,
"9.225 g" should be added.
Answer:
It would be compound.
Explanation:
It is this way because if it adds another proton it becomes more positive that nuetral, and if you add an electron it just makes the atom more dense. That is why the answer is compound. Hope this helped :)
