You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Answer:- 2.39 mL are required.
Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:
Where, 
and 
are molarities of concentrated and diluted solutions and 
and 
are their respective volumes.
= 1.10M
= 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)
= ?
= 525 mL
Let's plug in the given values in the formula:
So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.
Answer:
addition polymerization
Explanation:
In addition polymerization, the monomers are simply joined to each other to form a polymer having the same empirical formula as the monomer but of higher relative molecular mass. The monomers in addition polymerization are usually simple unsaturated molecules such as alkenes.
We can deduce the reaction to be an addition polymerization because of the the attachment of n to both the unsaturated monomer and the saturated polymer without the loss of any small molecule. If it was a condensation polymerization, there would have been an accompanying loss of a small molecule such as water.
There's a slight error in your equation. I think you were trying to present it like this:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Mole Ratio
O2 : H20
25 : 18
? moles : 18 moles
(18/18)×25 : 18 moles
25 moles : 18 moles
Final answer would be 25 moles of O2. :)
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