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.
"CH3CH2CH2CH2OH " is known by the name of "n-butanol" and "CH3CH(OH)CH3" is known by the name of "<span>Isopropyl alcohol". These two given products are basically alcohols. I hope that this is the answer that you were looking for and the answer has actually come to your desired help. Thanks for joining brainly and getting your questions solved.</span>
Answer:
temperature of the reaction vessel
Explanation:
temperature of the reaction vessel
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
First, let's count mole of 10 g Calcium Carbonate
mole = Mass / Molecular Mass
Calcium Carbonate = CaCO₃
Molecular Mass = Ar Ca + Ar C + (3 x Ar O)
Molecular Mass = 40 + 12 + (3 x 16)
Molecular Mass = 100
next
Mole of CaCO₃ = 10 gram / 100
Mole of CaCO₃ = 0,1 mol
then equal the reaction equation first
CaCO₃ + 2 HCl ==> CaCl₂ + CO₂ + H₂O (Equal)
To count the mass of carbon dioxide that produced we must know the mole of CO₂ first
we can count by coefficient comparison
mole CO₂ = 
x mole CaCO₃
mole CO₂ = (1/1) x 0,1 mole
mole CO₂ = 0,1 mole
so
Mass of CO₂ = mole CO₂ x Molecular Mass of CO₂
Mass of CO₂ = 0,1 mole x (12 + (2 x 16))
Mass of CO₂ = 0,1 mole x 44
Mass of CO₂ = 4,4 g
so, mass of carbon dioxide that's produced by 10 g of calcium carbonate on reaction with chloride acid is 4,4 g.
