Answer:
Carbonic acid could be formed.
Explanation:
Hello,
Based on her claim, it would be a really useful strategy to prevent global warming, nevertheless, there would be a problem if a increasing amount of carbon dioxide is not buried at the bottom of the ocean yet it flows freely along the sea and probably reacting with the water, causing carbonic acid to be formed and subsequently cutting back the sea's pH (increasing its acidity).
It would be useful, but a constant monitoring of the sea's pH must be needed because this could cause some species to be affected not only by the temperature but for the acid pH as well.
Best regards.
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
Answer:
A mushroom is a heterotroph.
Explanation:
Mushrooms are fungi, which are heterotrophs because they depend on other organisms for their food.
Answer:
Half will be green and half will be striped
Explanation:
Given green skin (G) is dominant to striped skin (g).
Cucumber 1 : Heterozygous (Gg)
Cucumber 2 : Homozygous striped (gg)
When Gg X gg :
G g
g Gg gg
g Gg gg
Half of the offspring will have Gg genotype so they will be green. Other half will have gg genotype so they will be striped.
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.
