From the equation q=mCΔT, set the q of copper = to q of water,
So --- mCΔT(copper)=mCΔT(water).
mass (Cu - copper) = 38g
mass (H2O - water) = 15g
C (H2O) = 4.184 J/g*C
ΔΤ (H2O) = 33-22 = 11*C
ΔΤ (Cu) = 33-80 = -47*C (the final temp is the same for both materials - thermal equilibrium)
C (Cu) = ?
So --- 38(-47)C[Cu]=15(4.184)(11)
--- C[Cu]=690.36/(-1786) = 0.3865 J/g*C, or 0.39 in 2 sig figs. (The negative goes away, because specific heats are usually positive)
Answer:
I think your answer is either b or c but I think b is more likely to be your answer
Answer:
Redox reaction and single displacement
Explanation:
This reaction is first of all a redox reaction. A redox reaction is a reaction that involves both oxidation and reduction. Oxidation involves increase in oxidation number while reduction involves decrease in oxidation number.
Copper (Cu) had an oxidation number of "0" as a reactant but had an oxidation number of "2+" in the product [Cu(NO₃)₂] hence oxidation occurred.
Nitrogen (N) had an oxidation number of "5+" in the reactant (HNO₃) but had an oxidation number of "4+" in the product (NO₂) hence reduction also occurred.
Also, from the reaction, it can be deduced that copper (Cu) displaced hydrogen (H) from the nitric acid (HNO₃) solution to form copper (II) nitrate [Cu(NO₃)₂]. It should be noted that copper can displace hydrogen because it is higher than hydrogen in the electrochemical series. Hence, this reaction can also be called a single displacement reaction. A single displacement reaction is a reaction in which an atom of an element replaces another atom in a compound (as seen in the equation given in the question).
I think it’s tissue I just did it
Hoffman Product is always a
less substituted alkene. In given scenario Structure
A (Given Below) was taken as a starting ch
iral Amine. For the sake of elimination the amine was taken tertiary so that on methylation it becomes a
good leaving group. This chiral amina (A) when treated with
Methyl Iodide gives a
quarternary amine which on treatment with Silver oxide yields less substituted Alkene (
B) as shown Below.
Alkene
B on Ozonolysis give two aldehydes i.e. Formaldehyde and Butyraldehyde.
