Reactives
-> Products
CuO
and water are products.
I
found this reaction which has CuO and water as products: decomposition of
Cu(OH)2.
Cu(OH)2
-> CuO + H2O
Stoichiometry calculus involve the mole
proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of
CuO and 1 mole of H2O are formed.
Considering
the molar masses:
Cu(OH)2
= 83.56 g/mol
CuO
= 79.545 g/mol
H2O
= 18.015 g/mol
Then:
When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.
You
should use that numbers in the rule of three:
79.545
g CuO __________18.015 g water
3.327
g CuO__________ x =3.327*18.015 /79.545 g water
x= 0.7535 g water
Biuret reagent. This is the answer
Methanoic acid has a molecular formula of HCOOH, when form an ester it's HCOO-, so the rest of the ester is -C4H9, a saturated 4 carbon chain. As shown in the attached diagram, there are 4 structural isomers, and the third isomer has 2 stereoisomers. So the answers is D:5.
Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s
Mostly and for what I would say is A
